# NCERT Solutions Class 11 Maths Chapter 3

## NCERT Solutions Class 11 Maths Chapter 3 – Trigonometric Functions

Class 11 and Class 12 Mathematics forms the fundamentals for many future subjects related to engineering, statistics, and artificial intelligence models. So it’s very important for students to get an in-depth understanding of all topics covered in Class 11 and Class 12 Mathematics.

Class 11 Chapter 3 Trigonometric functions is an essential topic as it prepares students for higher grades. It deals with the domain and range of trigonometric functions. In addition, it covers the sum and difference of two angles within the trigonometric functions.

Extramarks  Mathematics faculty experts have prepared detailed NCERT Mathematics solutions to help students with their Mathematics preparation. Students can access  NCERT Solutions Class 11 Mathematics Chapter 3  on the Extramarks’ website. It comprises a stepwise and detailed description of the various concepts covered in the chapter and also includes basic and advanced level questions. Students are advised to refer to NCERT Solutions Class 11 Mathematics Chapter 3   to perform better in the examination.

These NCERT Solutions notes are prepared by the Extramarks subject experts team with extensive teaching experience. Students of Class 11 can learn and revise essential points, definitions, and Q&A from the study material offered by NCERT Solutions for Class 11 Mathematics Chapter 3.

## Key Topics Covered In NCERT Solution for Class 11 Maths Chapter 3

The knowledge of Trigonometric Functions is an essential part of Class 11 Mathematics as it involves studying various relations with sides and angles. Trigonometric functions are used for basic geometric calculations and to explain numeric solutions.  Students must have a thorough knowledge of trigonometry as it  is a regular feature in many competitive exams. It is important to work out complex angles and dimensions in very little time,  therefore, students must apply test taking strategies  and they can refer to NCERT Solutions Class 11 Mathematics Chapter 3 to supplement their learning outcome and be satisfied with their preparation.

The key topics covered in NCERT Solutions Class 11 Mathematics Chapter 3 are as follows:

 Exercise Topics 3.1 Introduction 3.2 Trigonometric Equation and Angles 3.3 Trigonometric Functions and Ranges 3.4 Trigonometric Functions of Sum of Two Angles and Difference of Two Angles Others Miscellaneous

3.1 Introduction

In this section, the students are presented with basic trigonometric calculations and their use. It is expected that you calculate distances using these ratios. The degrees and the radians are two of the most commonly used measurement units for angles. Students studying mathematics and other subjects must understand the fundamentals of measuring angles with trigonometry. The questions provided in NCERT Solutions Class 11 Mathematics Chapter 3   are structured to help students understand core concepts of Trigonometry better.

3.2 Trigonometric Equation and Angles

Students will study measurements of angles in various units, degrees, and radians and their relationships—a solution to the numerical equation that requires converting measures of angles from one format to another. The exercise in Chapter 3 Mathematics Class 11 includes questions on graphs of trigonometric function graphs, domains, signs, and the range.

The exercise includes numerous examples and problems that will help you discover how a specific trigonometric relationship behaves in one or the other of four quadrants. Students can also examine graphs of various trigonometric ratios and how they perform under particular circumstances. Thus, Class 11 Mathematics NCERT Solutions Chapter 3 will benefit the students.

3.3 Trigonometric Functions and Ranges

Trigonometric functions connect the angle of a straight-angled triangle with the proportion of side lengths. Sin, Cos, and Tan are the three main functions. This practice consists of problems based on the trigonometric functions of sum and differences between two angles.

3.4 Sum and Difference of Two Angles

While solving problems with trigonometry, students will come across many problems and situations where they are required to calculate the trigonometric solutions for the sum of two angles or differences of two angles.

3.5 Miscellaneous

The questions in this exercise can be beneficial in revising the essential concepts that are related to Applications of Trigonometric functions.

The identities for the sum and difference are represented by the formulas below:

• sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
• cos(x+y) = cos(x)cos(y) – sin(x)sin(y)
• tan(x+y) = (tan x + tan y)/ (1−tan x • tan y)
• sin(x–y) = sin(x)cos(y) – cos(x)sin(y)
• cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
• tan(x−y) = (tan x–tan y)/ (1+tan x • tan y)

### NCERT Solutions Class 11 Maths Chapter 3: Exercise & Solutions

NCERT Solutions Class 11 Mathematics Chapter 3 explains the key concepts based on the trigonometric functions. Students can get access to the NCERT exercise and solutions prepared by Extramarks subject matter experts. It will help students to grasp the topic and essential concepts and also improve their problem-solving speed with accuracy. There are 61 questions in the NCERT Solutions Class 11 Mathematics Chapter 3 : Exercise and Solutions. Furthermore, it has 21 identity-based sums, 12 are intermediate level and 28 of high difficulty level.

Students can get  the exercise  NCERT Solutions Class 11 Mathematics Chapter 3 by clicking on the links  below.

• Class 11 Maths Chapter No. 3 Exercise 3.1 – 7 Questions
• Class 11 Maths Chapter No. 3 Exercise 3.2 – 10 Questions
• Class 11 Maths Chapter No. 3 Exercise 3.3 – 25 Questions
• Class 11 Maths Chapter No. 3 Exercise 3.4 – 9 Questions
• Class 11 Maths Chapter No. 3 Miscellaneous Exercise – 10 Questions

Students may access NCERT Solutions Class 11 Mathematics other chapters by clicking here. In addition, students can also explore NCERT solutions for other Classes below.

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10
• NCERT Solutions Class 11
• NCERT Solutions Class 12

### NCERT Exemplar Class 11 Maths

The NCERT Exemplar is designed for the students of Class 11 Mathematics under the guidance of CBSE. The Exemplar consists of multiple-choice questions, long-form type questions and short-form type questions. It helps students test their knowledge and preparation. So,  they are confident that they will get good marks in their annual exams. Furthermore, Exemplar solutions help to clear the toughest competitive exams. NCERT Exemplar covers all essential topics and concepts prescribed in the Class 11 Mathematics Chapter 3 syllabus.

To score better in the exam, the students can also refer to the NCERT Solutions Class 11 Mathematics Chapter 3 prepared by the Extramarks experienced faculty. Students who have clear theoretical concepts or a strong foundational base can start practising with Exemplar. Both solutions booklet and  Exemplar’s notes are available on the Extramarks’ website.

### Key Features of NCERT Solutions Class 11 Maths Chapter 3

NCERT Solutions Class 11 Mathematics Chapter 3 promotes core fundamentals to have a firm grip on the topics. The solution set covers topics such as trigonometric functions, equations and formulas.

The key features of Extramarks NCERT Solutions for Class 11 Mathematics Chapter 3 include:

• The solutions will help the students to get a quick review of the chapter ahead of the exam.
• It covers essential questions for the board exams and the competitive examinations.
• With the help of NCERT Solutions Class 11 Mathematics Chapter 3, the students can develop a strong foundation in Trigonometric Functions.
• It discusses crucial topics such as trigonometric ratios of acute angles, general solutions, sine, cosine, and other trigonometry formulas. Students can learn the trigonometric ratios and definitions of right angles and hypotenuse sides.

Q.1 Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) – 47°30′
(iii) 240°
(iv) 520°

Ans.

$\begin{array}{l}\text{Since, 1°}=\frac{\text{π}}{180\text{°}}\\ \left(\text{i}\right)\text{ }25\text{°}=25\text{°}×\frac{\text{π}}{180\text{°}}\\ \text{ }=\frac{5\text{π}}{36\text{°}}\text{ radians}\\ \left(\text{ii}\right)\text{ }-47\text{°}30‘=-47.5\text{° }\left[\because \text{ }30‘=0.5\text{°}\right]\\ \text{ }=-\text{\hspace{0.17em}}47.5\text{°}×\frac{\text{π}}{180\text{°}}\text{ radians}\\ \text{\hspace{0.17em} }=-\frac{19\text{π}}{72}\text{ radians}\\ \left(\text{iii}\right)\text{ }240\text{°}=240\text{°}×\frac{\text{π}}{180\text{°}}\text{ radians}\\ \text{ }=\frac{4\text{π}}{3}\text{ radians}\\ \left(\text{iv}\right)\text{ }520\text{°}=520\text{°}×\frac{\text{π}}{180\text{°}}\text{ radians}\\ \text{ }=\frac{26\text{π}}{9}\text{ radians}\end{array}$

Q.2

$\begin{array}{l}\text{Find the degree measures corresponding to the following}\\ \text{radian measures }\left(\text{Use π}=\frac{22}{7}\right).\\ \left(\text{i}\right)\text{ }\frac{11}{16}\\ \left(\text{ii}\right)\text{ }-4\\ \left(\text{iii}\right)\text{ }\frac{5\text{π}}{3}\\ \left(\text{iv}\right)\text{ }\frac{7\text{π}}{6}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{1 radian}=\frac{180\mathrm{°}}{\mathrm{\pi }}\\ \left(\mathrm{i}\right)\frac{11}{16}\text{\hspace{0.17em}\hspace{0.17em}radians}=\frac{11}{16}×\frac{180\mathrm{°}}{\mathrm{\pi }}\\ =\frac{11}{4}×\frac{45\mathrm{°}}{\mathrm{\pi }}\\ =\frac{11}{4}×\frac{45\mathrm{°}×7}{22}\\ =\frac{1}{4}×\frac{45\mathrm{°}×7}{2}\\ =\left(39\frac{3}{8}\right)°\\ =39\mathrm{°}+\frac{3}{8}×60‘\\ =39\mathrm{°}+22‘+\frac{1}{2}‘\\ =39\mathrm{°}+22‘+\frac{1}{2}×60‘‘\\ =39\mathrm{°}22‘30‘‘\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)-4\text{\hspace{0.17em}\hspace{0.17em}radians}=-4×\frac{180\mathrm{°}}{\mathrm{\pi }}\\ =-4×\frac{180\mathrm{°}×7}{22}\\ =-\left(229\frac{1}{11}\right)°\\ =-\left(229\mathrm{°}+\frac{1}{11}×60‘\right)\\ =-\left(229\mathrm{°}+5‘+\frac{5}{11}×60‘‘\right)\\ =-\left(229\mathrm{°}+5‘+27‘‘\right)\\ =-229\mathrm{°}5‘27‘‘\left(\mathrm{Approx}.\right)\\ \left(\mathrm{iii}\right)\frac{5\mathrm{\pi }}{3}\text{\hspace{0.17em}\hspace{0.17em}radians}=\frac{5\mathrm{\pi }}{3}×\frac{180\mathrm{°}}{\mathrm{\pi }}\\ =\frac{5}{3}×180\mathrm{°}\\ =300\mathrm{°}\\ \left(\mathrm{iv}\right)\frac{7\mathrm{\pi }}{6}\text{\hspace{0.17em}\hspace{0.17em}radians}=\frac{7\mathrm{\pi }}{6}×\frac{180\mathrm{°}}{\mathrm{\pi }}\\ =\frac{7}{6}×180\mathrm{°}\\ =210\mathrm{°}\end{array}$

Q.3 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Ans.

$\begin{array}{l}\mathrm{Number}\text{of revolutions made by a wheel in 1 minute}=360\\ \mathrm{Number}\text{of revolutions made by a wheel in 1 second}=\frac{360}{60}\\ \text{ }=6\\ \mathrm{Angle}\text{covered in 1 revolution by wheel}=2\text{π }\mathrm{Radians}\\ \mathrm{Angle}\text{covered in 6 revolutions by wheel}=6×2\text{π}\\ \text{ }=\text{ }12\text{π }\mathrm{Radians}\\ \mathrm{Thus},\text{wheel turns 12π radians in 1 second.}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{degree}\mathrm{measure}\mathrm{of}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{at}\mathrm{the}\\ \mathrm{centre}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{of}\mathrm{radius}100\mathrm{cm}\mathrm{by}\mathrm{an}\mathrm{arc}\mathrm{of}\mathrm{length}22\mathrm{cm}\\ \left(\mathrm{Use}\mathrm{\pi }=\frac{22}{7}\right).\end{array}$

Ans.

$\begin{array}{l}\mathrm{Radius}\text{of circle}=100\text{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Length​ of arc}=22\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{Angle}\left(\mathrm{\theta }\right)\text{subtended at the centre by an arc}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{Arc}}{\mathrm{Radius}}\text{\hspace{0.17em}}\mathrm{radians}\\ \text{\hspace{0.17em}}\mathrm{\theta }\text{\hspace{0.17em}}=\frac{22}{100}\text{\hspace{0.17em}}\mathrm{radians}\\ \text{\hspace{0.17em}}\mathrm{\theta }\text{\hspace{0.17em}}=\frac{22}{100}\text{\hspace{0.17em}}×\frac{180\mathrm{°}}{\mathrm{\pi }}\left[\because 1\text{\hspace{0.17em}}\mathrm{radian}=\frac{180\mathrm{°}}{\mathrm{\pi }}\right]\\ \text{\hspace{0.17em}}\mathrm{\theta }\text{\hspace{0.17em}}=\frac{22}{100}\text{\hspace{0.17em}}×\frac{180\mathrm{°}×7}{22}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1.8\mathrm{°}×7\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12.6\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12\mathrm{°}+0.6×60‘\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12\mathrm{°}+36‘\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12\mathrm{°}36‘\\ \text{Thus, angle subtended by an arc of 22 cm at the centre is 12°36′ .}\end{array}$

Q.5 In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

Ans.

$\begin{array}{l}\mathrm{Diameter}\text{of a circle}=40\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Radius}\text{​ of a circle}=\frac{40}{2}\text{\hspace{0.17em}}\mathrm{cm}\\ =20\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of chord}=20\text{\hspace{0.17em}}\mathrm{cm}\end{array}$

$\begin{array}{l}\mathrm{Since},\text{​}\mathrm{\Delta }\text{OAB is equilateral triangle.}\\ \mathrm{So},\text{​ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{AOB}=60\mathrm{°}\\ =\frac{\mathrm{\pi }}{3}\mathrm{radians}\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}angle}=\frac{\mathrm{Length}\text{of arc}}{\mathrm{Radius}}\\ \frac{\mathrm{\pi }}{3}\mathrm{Radian}=\frac{\mathrm{Length}\text{of arc AB}}{20\mathrm{cm}}\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{Length}\text{of arc AB}=20\mathrm{cm}×\frac{\mathrm{\pi }}{3}\\ =\frac{20\mathrm{\pi }}{3}\mathrm{cm}\end{array}$

Q.6 If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Ans.

$\begin{array}{l}\mathrm{L}{\text{et radii of circles be r}}_{\text{1}}{\text{and r}}_{\text{2}}\text{respectively.}\\ \text{Angle subscribed by arc in first circle}=60\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}=60\mathrm{°}×\frac{\mathrm{\pi }}{180\mathrm{°}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{radians}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{\pi }}{3}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{radians}\\ \text{Angle subscribed by arc in second circle}\\ \text{\hspace{0.17em}\hspace{0.17em}}=75\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}=75\mathrm{°}×\frac{\mathrm{\pi }}{180\mathrm{°}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{radians}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{5\mathrm{\pi }}{12}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{radians}\\ \mathrm{Since},\text{length of both arcs is same.}\\ \text{So,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Length of arc of first circle}=\text{Length of arc of}\\ \text{second circle}\\ \frac{\mathrm{\pi }}{3}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{radian}×{\mathrm{r}}_{1}=\frac{5\mathrm{\pi }}{12}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{radian}×{\mathrm{r}}_{2}\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{r}}_{1}}{{\mathrm{r}}_{2}}=\frac{\left(\frac{5\mathrm{\pi }}{12}\right)}{\left(\frac{\mathrm{\pi }}{3}\right)}\\ =\frac{5}{4}\\ ⇒{\mathrm{r}}_{1}:{\mathrm{r}}_{2}=5:4\\ \mathrm{Thus},\text{the ratio of radii of circles is 5:4.}\end{array}$

Q.7 Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm

Ans.

$\begin{array}{l}\mathrm{Length}\text{of pendulum}=\text{75 cm}\\ \left(\text{i}\right)\text{ }\mathrm{Length}\text{of arc described by pendulum}=10\text{ }\mathrm{cm}\\ \mathrm{Angle}\text{swept by pendulum}=\frac{10}{75}\text{ }\mathrm{Rad}.\left(\because \text{ θ}=\frac{\mathrm{arc}}{\text{r}}\right)\\ \text{ }=\frac{2}{15}\text{ }\mathrm{radians}\\ \left(\mathrm{ii}\right)\text{ }\mathrm{Length}\text{of arc described by pendulum}=15\text{ }\mathrm{cm}\\ \mathrm{Angle}\text{swept by pendulum}=\frac{15}{75}\text{ }\mathrm{Rad}.\left(\because \text{ θ}=\frac{\mathrm{arc}}{\text{r}}\right)\\ \text{ }=\frac{1}{5}\text{ }\mathrm{radians}\\ \left(\mathrm{iii}\right)\text{ }\mathrm{Length}\text{of arc described by pendulum}\\ \text{ }=21\text{ }\mathrm{cm}\\ \mathrm{Angle}\text{swept by pendulum}=\frac{21}{75}\text{ }\mathrm{Rad}.\left(\because \text{ θ}=\frac{\mathrm{arc}}{\text{r}}\right)\\ \text{ }=\frac{7}{25}\text{ }\mathrm{radians}\end{array}$

Q.8 Find the values of other five trigonometric functions in Exercises 1 to 5.

$1.\text{ }\mathrm{cos}\text{x}=-\frac{1}{2},\text{x}\mathrm{lies}\mathrm{in}\mathrm{third}\mathrm{quadrant}.$ $2.\text{ }\mathrm{sin}\text{x}=\frac{3}{5},\text{x}\mathrm{lies}\mathrm{in}\mathrm{second}\mathrm{quadrant}.$ $3.\text{ }\mathrm{cot}\text{x}=\frac{3}{4},\text{x}\mathrm{lies}\mathrm{in}\mathrm{third}\mathrm{quadrant}.$ $4.\text{ }\mathrm{sec}\text{x}=\frac{13}{5},\text{x}\mathrm{lies}\mathrm{in}\mathrm{fourth}\mathrm{quadrant}.$ $5.\text{ }\mathrm{tan}\text{x}=-\frac{5}{12},\text{x}\mathrm{lies}\mathrm{in}\mathrm{second}\mathrm{quadrant}.$

Ans.

$\begin{array}{l}1.\\ \mathrm{secx}=\frac{1}{\mathrm{cosx}}\\ \text{ }=\frac{1}{-\frac{1}{2}}\\ \text{ }=-2\left(\mathrm{x}\text{ }\mathrm{lies}\text{ }\mathrm{in}\text{ }\mathrm{third}\text{ }\mathrm{quadrant}.\right)\\ {\mathrm{sin}}^{2}\text{x}+{\mathrm{cos}}^{2}\text{x}=1\\ \text{ }\mathrm{sinx}=\sqrt{1-{\mathrm{cos}}^{2}\text{x}}\\ \text{ }=\sqrt{1-{\left(-\frac{1}{2}\right)}^{2}}\\ \text{ }\mathrm{sinx}=\sqrt{\frac{3}{4}}\\ \mathrm{sinx}=-\frac{\sqrt{3}}{2}\text{ }\left(\because \text{ x }\mathrm{lies}\text{in third quadrant.}\right)\\ \mathrm{cosecx}=\frac{1}{\mathrm{sinx}}\\ \text{ }=\frac{1}{\left(-\frac{\sqrt{3}}{2}\right)}\\ \text{ }=-\frac{2}{\sqrt{3}}\\ \text{ }\mathrm{tanx}=\frac{\mathrm{sinx}}{\mathrm{cosx}}\\ \text{ }=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\\ \text{ }=\sqrt{3}\left(\because \text{x }\mathrm{lies}\text{in third quadrant.}\right)\\ \text{ }\mathrm{cotx}=\frac{1}{\mathrm{tanx}}\\ \text{ }=\frac{1}{\sqrt{3}}\left(\because \text{x }\mathrm{lies}\text{in third quadrant.}\right)\\ \mathrm{Thus},\text{sinx}=-\frac{\sqrt{3}}{2},\text{}\mathrm{cosecx}=-\frac{2}{\sqrt{3}},\text{ }\mathrm{secx}=-2\text{,tanx}=\sqrt{3}\text{ }\mathrm{and}\text{ }\mathrm{cotx}=\frac{1}{\sqrt{3}}.\end{array}$ $\begin{array}{l}\text{2.}\mathrm{sinx}=3/5\\ \mathrm{cosecx}=1/\mathrm{sinx}\\ \text{ }=\frac{1}{\left(\frac{3}{5}\right)}\\ \mathrm{cosec}\text{ }\mathrm{x}=\frac{5}{3}\\ {\mathrm{sin}}^{2}\text{x}+{\mathrm{cos}}^{2}\text{x}=1\\ \mathrm{cos}\text{ }\mathrm{x}=±\sqrt{1-{\mathrm{sin}}^{2}\text{x}}\\ \text{ }=±\sqrt{1-{\left(\frac{3}{5}\right)}^{2}}\\ \mathrm{cos}\text{ }\mathrm{x}=±\sqrt{\frac{16}{25}}\\ \text{ }=-\frac{4}{5}\text{ }\left[\because \text{ x lies in II quadrant.}\right]\\ \mathrm{sec}\text{ }\mathrm{x}=\frac{1}{\mathrm{cos}\text{ }\mathrm{x}}\\ \text{ }=\frac{1}{-\left(\frac{4}{5}\right)}\\ \text{ }=-\frac{5}{4}\\ \mathrm{tan}\text{ }\mathrm{x}=\frac{\mathrm{sin}\text{ }\mathrm{x}}{\mathrm{cos}\text{ }\mathrm{x}}\\ \mathrm{tan}\text{ }\mathrm{x}=\frac{\left(\frac{3}{5}\right)}{-\left(\frac{4}{5}\right)}\\ \text{ }=-\frac{3}{4}\text{ }\left[\because \text{ x lies in II quadrant.}\right]\\ \mathrm{cot}\text{ }\mathrm{x}=\frac{1}{\mathrm{tan}\text{ }\mathrm{x}}\\ \text{ }=\frac{1}{\left(-\frac{3}{4}\right)}\\ \text{ }=-\frac{4}{3}\text{ }\left[\because \text{ x lies in II quadrant.}\right]\\ \mathrm{Thus},\text{cos x}=-\frac{4}{5},\text{ }\mathrm{sec}\text{ }\mathrm{x}=-\frac{5}{4}\text{ ,}\mathrm{cosec}\text{ }\mathrm{x}=\frac{5}{3},\text{tan x}=-\frac{3}{4}\\ \mathrm{and}\text{ }\mathrm{cot}\text{ }\mathrm{x}=-\frac{4}{3}.\end{array}$ $\begin{array}{l}3.\\ \mathrm{tan}\text{ }\mathrm{x}=\frac{1}{\mathrm{cot}\text{ }\mathrm{x}}\\ \text{ }=\frac{1}{\left(\frac{3}{4}\right)}\\ \text{ }=\frac{4}{3}\\ {\mathrm{cosec}}^{2}\text{x}=1+{\mathrm{cot}}^{2}\text{x}\\ \text{ }=1+{\left(\frac{3}{4}\right)}^{2}\\ \text{ }=1+\frac{9}{16}\\ \mathrm{cosec}\text{ }\mathrm{x}=±\sqrt{\frac{25}{16}}\\ \text{ }=-\frac{5}{4}\text{ }\left[\text{x lies in third quadrant.}\right]\\ \text{ }\mathrm{sin}\text{ }\mathrm{x}=\frac{1}{\mathrm{cosec}\text{ }\mathrm{x}}\\ \text{ }=\frac{1}{-\frac{5}{4}}\\ \text{ }=-\frac{4}{5}\\ \text{ }\mathrm{sec}\text{ }\mathrm{x}=1+{\mathrm{tan}}^{2}\text{x}\\ \text{ }=1+{\left(\frac{4}{3}\right)}^{2}\\ \text{ }=1+\frac{16}{9}\\ \text{ }\mathrm{sec}\text{ }\mathrm{x}=±\sqrt{\frac{25}{9}}\\ \text{ }=-\frac{5}{3}\left[\text{x lies in third quadrant.}\right]\\ \text{ cos x}=-\frac{3}{5}\left[\text{x lies in third quadrant.}\right]\\ \mathrm{Thus},\text{ }\mathrm{sin}\text{ }\mathrm{x}=-\frac{4}{5},\text{ }\mathrm{cosec}\text{ }\mathrm{x}=-\frac{5}{4},\mathrm{cos}\text{ }\mathrm{x}=-\frac{3}{5},\text{ }\mathrm{sec}\text{ }\mathrm{x}=-\frac{5}{3}\text{ }\mathrm{and}\text{ }\mathrm{tan}\text{ }\mathrm{x}=\frac{4}{3}.\end{array}$ $\begin{array}{l}4.\\ \mathrm{cos}\text{x}=\frac{1}{\mathrm{sec}\text{x}}\\ \text{ }=\frac{1}{\left(\frac{13}{5}\right)}\\ \text{ }=\frac{5}{13}\text{ }\left[\because \text{ }\text{x lies in fourth quadrant.}\right]\\ \mathrm{sin}\text{x}=\sqrt{1-{\mathrm{cos}}^{2}\text{x}}\\ \text{ }=±\sqrt{1-{\left(\frac{5}{13}\right)}^{2}}\\ \text{ }=-\frac{12}{13}\text{ }\left[\because \text{ }\text{x lies in fourth quadrant.}\right]\\ \mathrm{cosec}\text{x}=\frac{1}{\mathrm{sin}\text{x}}\\ \text{ }=\frac{1}{\left(-\frac{12}{13}\right)}\\ \text{ }=-\frac{13}{12}\\ \text{ }\mathrm{tan}\text{x}=\frac{\mathrm{sin}\text{x}}{\mathrm{cos}\text{x}}\end{array}$ $\begin{array}{l}=\frac{-\left(\frac{12}{13}\right)}{\left(\frac{5}{13}\right)}\\ =-\frac{12}{5}\left[\because \mathrm{x}\text{lies in fourth quadrant.}\right]\\ \mathrm{cot}\mathrm{x}=\frac{1}{\mathrm{tanx}}\\ =\frac{1}{-\left(\frac{12}{5}\right)}\\ =-\frac{5}{12}\\ \mathrm{Thus},\text{}\mathrm{sin}\mathrm{x}=-\frac{12}{13},\text{\hspace{0.17em}}\mathrm{cos}\mathrm{x}=\frac{5}{13},\text{\hspace{0.17em}}\mathrm{tan}\mathrm{x}=-\frac{12}{5},\text{\hspace{0.17em}}\mathrm{cot}\mathrm{x}=-\frac{5}{12}\\ \mathrm{and}\text{}\mathrm{cosec}\mathrm{x}=-\frac{13}{12}.\end{array}$ $\begin{array}{l}5.\\ \mathrm{cot}\text{ x}=\frac{1}{\left(-\frac{5}{12}\right)}\left[\because \text{ }\mathrm{cot}\text{ }\text{x}=\frac{1}{\mathrm{tan}\text{ x}}\right]\\ \text{ }=-\frac{12}{5}\\ {\mathrm{sec}}^{2}\text{x}=1+{\mathrm{tan}}^{2}\text{x}\\ \text{ }=1+{\left(-\frac{5}{12}\right)}^{2}\\ \text{ }=1+\frac{25}{144}\\ \mathrm{sec}\text{ }\mathrm{x}=±\sqrt{\frac{169}{144}}\\ \text{ }=-\frac{13}{12}\text{ }\left[\because \text{ x lies in second quadrant.}\right]\\ \mathrm{cos}\text{ }\mathrm{x}=\frac{1}{\mathrm{secx}}\\ \text{ }=\frac{1}{-\left(\frac{13}{12}\right)}\\ \text{ }=-\frac{12}{13}\text{ }\left[\because \text{ x lies in second quadrant.}\right]\\ {\mathrm{cosec}}^{2}\text{x}=1+{\mathrm{cot}}^{2}\text{x}\\ \text{ }=1+{\left(-\frac{12}{5}\right)}^{2}\\ \text{ }=1+\frac{144}{25}\\ \mathrm{cosec}\text{ }\mathrm{x}=±\sqrt{\frac{169}{25}}\\ \text{ }=\frac{13}{5}\text{ }\left[\because \text{ x lies in second quadrant.}\right]\\ \text{ }\mathrm{sin}\text{ }\mathrm{x}=\frac{1}{\mathrm{cosec}\text{ }\mathrm{x}}\\ \text{ }=\frac{1}{\left(\frac{13}{5}\right)}\\ \mathrm{sin}\text{ }\mathrm{x}=\frac{5}{13}\text{ }\left[\because \text{ x lies in second quadrant.}\right]\\ \mathrm{Thus},\text{}\mathrm{sin}\text{ }\mathrm{x}=\frac{5}{13},\text{ }\mathrm{cos}\text{ }\mathrm{x}=-\frac{12}{13},\text{ }\mathrm{cot}\text{ }\mathrm{x}=-\frac{12}{5},\text{ }\mathrm{sec}\text{ }\mathrm{x}=-\frac{13}{12}\\ \mathrm{and}\text{ }\mathrm{cosec}\text{ }\mathrm{x}=\frac{13}{5}.\end{array}$

Q.9 Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°
7. cosec (– 1410°)

$8.\mathrm{tan}\frac{19\mathrm{\pi }}{3}$

$9.\mathrm{sin}\left(-\frac{11\mathrm{\pi }}{3}\right)$

$10.\mathrm{cot}\left(-\frac{15\mathrm{\pi }}{4}\right)$

Ans.

$\begin{array}{l}6.\\ \text{sin}\left(\text{765°}\right)\text{}=\text{sin}\left(\text{2 x 36}0\text{°}+\text{45°}\right)\\ \text{ }=\text{sin 45°}\left[\because \mathrm{sin}\left(\text{n}×360\text{°}+\text{θ}\right)=\text{sinθ}\right]\\ \text{ }=\frac{1}{\sqrt{2}}\end{array}$ $\begin{array}{l}7.\\ \text{​}\begin{array}{l}\text{cosec}\left(–\text{141}0\text{°}\right)=-\mathrm{cosec}\text{141}0\text{°\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{cosec}\left(-\text{θ}\right)=-\mathrm{cosec\theta }\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{cosec}\left(4×360\text{°}-30\text{°}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\left(-\mathrm{cosec}30\text{°}\right)\left[\mathrm{cosec}\left(\text{n}×360\text{°}-\text{θ}\right)=-\mathrm{cosec\theta }\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\end{array}\end{array}$ $\begin{array}{l}8.\\ \mathrm{tan}\frac{19\text{π}}{3}=\mathrm{tan}\left(6\text{π}+\frac{\text{π}}{3}\right)\end{array}$ $\begin{array}{l}\mathrm{tan}\frac{19\mathrm{\pi }}{3}=\mathrm{tan}\frac{\mathrm{\pi }}{3}\left[\because \mathrm{tan}\left(2\mathrm{n\pi }+\mathrm{\theta }\right)=\mathrm{tan\theta }\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{3}\end{array}$ $\begin{array}{l}9.\\ \text{​}\begin{array}{l}\mathrm{sin}\left(-\frac{11\text{π}}{3}\right)=-\mathrm{sin}\left(\frac{11\text{π}}{3}\right)\left[\because \mathrm{sin}\left(-\text{x}\right)=-\mathrm{sin}\text{x}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{sin}\left(4\text{π}-\frac{\text{π}}{3}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\left(-\mathrm{sin}\frac{\text{π}}{3}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \mathrm{sin}\left(2\mathrm{n\pi }-\text{x}\right)=-\mathrm{sin}\text{x}\right]\\ \mathrm{sin}\left(-\frac{11\text{π}}{3}\right)=\frac{\sqrt{3}}{2}\end{array}\end{array}$ $\begin{array}{l}10.\\ \text{​}\begin{array}{l}\mathrm{cot}\left(-\frac{15\text{π}}{4}\right)=-\mathrm{cot}\left(\frac{15\text{π}}{4}\right)\left[\because \mathrm{cot}\left(-\text{x}\right)=-\mathrm{cot}\text{x}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{cot}\left(4\text{π}-\frac{\text{π}}{4}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\left(-\mathrm{cot}\frac{\text{π}}{4}\right)\left[\because \mathrm{cot}\left(2\mathrm{n\pi }-\text{x}\right)=-\mathrm{cot}\text{x}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1\end{array}\end{array}$

Q.10

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\mathrm{sin}}^{2}\frac{\mathrm{\pi }}{6}+{\mathrm{cos}}^{2}\frac{\mathrm{\pi }}{3}-{\mathrm{tan}}^{2}\frac{\mathrm{\pi }}{4}=-\frac{1}{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \text{\hspace{0.17em}}{\mathrm{sin}}^{2}\frac{\mathrm{\pi }}{6}+{\mathrm{cos}}^{2}\frac{\mathrm{\pi }}{3}-{\mathrm{tan}}^{2}\frac{\mathrm{\pi }}{4}={\left(\frac{1}{2}\right)}^{2}+{\left(\frac{1}{2}\right)}^{2}-{\left(1\right)}^{2}\\ =\frac{1}{4}+\frac{1}{4}-1\\ =\frac{1+1-4}{4}\\ =-\frac{2}{4}\\ =-\frac{1}{2}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ 2{\mathrm{sin}}^{2}\left(\frac{\mathrm{\pi }}{6}\right)+{\mathrm{cosec}}^{2}\left(\frac{7\mathrm{\pi }}{6}\right)\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{3}{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ 2{\mathrm{sin}}^{2}\frac{\mathrm{\pi }}{6}+{\mathrm{cosec}}^{2}\frac{7\mathrm{\pi }}{6}\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\frac{\mathrm{\pi }}{3}=2{\left(\frac{1}{2}\right)}^{2}+{\mathrm{cosec}}^{2}\left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)×{\left(\frac{1}{2}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2×\frac{1}{4}+\frac{1}{4}×{\mathrm{cosec}}^{2}\left(\frac{\mathrm{\pi }}{6}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}+\frac{1}{4}×{\left(2\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{2}=\text{\hspace{0.17em}}\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.12

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\mathrm{cot}}^{2}\frac{\mathrm{\pi }}{6}+\mathrm{cosec}\frac{5\mathrm{\pi }}{3}+3\text{\hspace{0.17em}}{\mathrm{tan}}^{2}\frac{\mathrm{\pi }}{6}=6\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}:\\ {\mathrm{cot}}^{2}\frac{\mathrm{\pi }}{6}+\mathrm{cosec}\frac{5\mathrm{\pi }}{6}+3\text{\hspace{0.17em}}{\mathrm{tan}}^{2}\frac{\mathrm{\pi }}{6}={\left(\sqrt{3}\right)}^{2}+\mathrm{cosec}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)+3{\left(\frac{1}{\sqrt{3}}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3+\mathrm{cosec}\left(\frac{\mathrm{\pi }}{6}\right)+3×\frac{1}{3}\\ \left[\because \mathrm{cosec}\left(\mathrm{\pi }-\mathrm{x}\right)=\mathrm{cosec}\mathrm{x}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3+2+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.13

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ 2{\mathrm{sin}}^{2}\left(\frac{3\mathrm{\pi }}{4}\right)+2{\mathrm{cos}}^{2}\left(\frac{\mathrm{\pi }}{4}\right)+2{\mathrm{sec}}^{2}\left(\frac{\mathrm{\pi }}{3}\right)=10\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ 2{\mathrm{sin}}^{2}\left(\frac{3\mathrm{\pi }}{4}\right)+2{\mathrm{cos}}^{2}\left(\frac{\mathrm{\pi }}{4}\right)+2{\mathrm{sec}}^{2}\left(\frac{\mathrm{\pi }}{3}\right)=2{\mathrm{sin}}^{2}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)+2{\left(\frac{1}{\sqrt{2}}\right)}^{2}+2{\left(2\right)}^{2}\end{array}$ $\begin{array}{l}=2{\mathrm{sin}}^{2}\left(\frac{\mathrm{\pi }}{4}\right)+2\left(\frac{1}{2}\right)+8\\ =2{\left(\frac{1}{\sqrt{2}}\right)}^{2}+1+8\\ =2\left(\frac{1}{2}\right)+1+8\\ =1+1+8\\ =10=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.14 Find the value of: (i) sin75° (ii) tan15°

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{sin}75\mathrm{°}=\mathrm{sin}\left(45\mathrm{°}+30\mathrm{°}\right)\\ =\mathrm{sin}45\mathrm{°}\mathrm{cos}30\mathrm{°}+\mathrm{sin}30\mathrm{°}\mathrm{cos}45\mathrm{°}\\ \left[\because \mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{y}+\mathrm{sin}\mathrm{y}\mathrm{cos}\mathrm{x}\right]\\ =\frac{1}{\sqrt{2}}×\frac{\sqrt{3}}{2}+\frac{1}{2}×\frac{1}{\sqrt{2}}\\ =\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}\\ =\frac{\sqrt{3}+1}{2\sqrt{2}}\\ \left(\mathrm{ii}\right)\mathrm{tan}15\mathrm{°}=\mathrm{tan}\left(45\mathrm{°}-30\mathrm{°}\right)\\ =\frac{\mathrm{tan}45\mathrm{°}-\mathrm{tan}30\mathrm{°}}{1+\mathrm{tan}45\mathrm{°}\mathrm{tan}30\mathrm{°}}\\ =\frac{1-\frac{1}{\sqrt{3}}}{1+1×\frac{1}{\sqrt{3}}}\end{array}$ $\begin{array}{l}=\frac{\sqrt{3}-1}{\sqrt{3}+1}×\frac{\sqrt{3}-1}{\sqrt{3}-1}\\ =\frac{{\left(\sqrt{3}-1\right)}^{2}}{3-1}\\ =\frac{3-2\sqrt{3}+1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan}\text{}15\mathrm{°}=\frac{4-2\sqrt{3}}{2}\\ =2-\sqrt{3}\end{array}$

Q.15

$\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}-\mathrm{y}\right)-\mathrm{sin}\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)\mathrm{sin}\left(\frac{\mathrm{\pi }}{4}-\mathrm{y}\right)=\mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)$

Ans.

$\begin{array}{l}\mathrm{Since}, \mathrm{cosA}\mathrm{cosB}-\mathrm{sinA}\mathrm{sinB}=\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\\ \mathrm{cos}\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}-\mathrm{y}\right)-\mathrm{sin}\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)\mathrm{sin}\left(\frac{\mathrm{\pi }}{4}-\mathrm{y}\right)\\ =\mathrm{cos}\left\{\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)+\left(\frac{\mathrm{\pi }}{4}-\mathrm{y}\right)\right\}\\ =\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}-\mathrm{y}\right)\\ =\mathrm{cos}\left\{\frac{\mathrm{\pi }}{2}-\left(\mathrm{x}+\mathrm{y}\right)\right\}\\ =\mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)\left[\because \mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)=\mathrm{sin}\mathrm{\theta }\right]\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Prove}\mathrm{the}\mathrm{following}:\\ \frac{\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}+\mathrm{x}\right)}{\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)}={\left(\frac{1+\mathrm{tan}\mathrm{x}}{1-\mathrm{tan}\mathrm{x}}\right)}^{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}:\\ \frac{\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}+\mathrm{x}\right)}{\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)}=\text{}\frac{\left(\frac{\mathrm{tan}\frac{\mathrm{\pi }}{4}+\mathrm{tan}\mathrm{x}}{1-\mathrm{tan}\frac{\mathrm{\pi }}{4}\mathrm{tan}\mathrm{x}}\right)}{\left(\frac{\mathrm{tan}\frac{\mathrm{\pi }}{4}-\mathrm{tan}\mathrm{x}}{1+\mathrm{tan}\frac{\mathrm{\pi }}{4}\mathrm{tan}\mathrm{x}}\right)}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\because \mathrm{tan}\left(\mathrm{A}+\mathrm{B}\right)=\frac{\mathrm{tan}\mathrm{A}+\mathrm{tan}\mathrm{B}}{1-\mathrm{tan}\mathrm{A}\mathrm{tan}\mathrm{B}}\\ \mathrm{tan}\left(\mathrm{A}-\mathrm{B}\right)=\frac{\mathrm{tan}\mathrm{A}-\mathrm{tan}\mathrm{B}}{1+\mathrm{tan}\mathrm{A}\mathrm{tan}\mathrm{B}}\end{array}\right]\\ =\text{}\frac{\left(\frac{1+\mathrm{tan}\mathrm{x}}{1-1.\mathrm{tan}\mathrm{x}}\right)}{\left(\frac{1-\mathrm{tan}\mathrm{x}}{1+1.\mathrm{tan}\mathrm{x}}\right)}\\ =\left(\frac{1+\mathrm{tan}\mathrm{x}}{1-\mathrm{tan}\mathrm{x}}\right)×\left(\frac{1+\mathrm{tan}\mathrm{x}}{1-\mathrm{tan}\mathrm{x}}\right)\\ ={\left(\frac{1+\mathrm{tan}\mathrm{x}}{1-\mathrm{tan}\mathrm{x}}\right)}^{2}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.17

$\begin{array}{l}\mathrm{Prove}\mathrm{the}\mathrm{following}:\\ \frac{\mathrm{cos}\left(\mathrm{\pi }+\mathrm{x}\right)\mathrm{cos}\left(-\mathrm{x}\right)}{\mathrm{sin}\left(\mathrm{\pi }-\mathrm{x}\right)\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}+\mathrm{x}\right)}={\mathrm{cot}}^{2}\mathrm{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\mathrm{cos}\left(\mathrm{\pi }+\mathrm{x}\right)\mathrm{cos}\left(-\mathrm{x}\right)}{\mathrm{sin}\left(\mathrm{\pi }-\mathrm{x}\right)\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}+\mathrm{x}\right)}=\frac{-\mathrm{cos}\mathrm{x}\mathrm{cos}\mathrm{x}}{\mathrm{sin}\mathrm{x}×-\mathrm{sin}\mathrm{x}}\left[\begin{array}{l}\because \mathrm{cos}\left(\mathrm{\pi }+\mathrm{x}\right)=-\mathrm{cosx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos}\left(-\mathrm{x}\right)=\mathrm{cosx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{sin}\left(\mathrm{\pi }-\mathrm{x}\right)=\mathrm{sinx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}+\mathrm{x}\right)=-\mathrm{sinx}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{cot}}^{2}\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.18

$\mathrm{cos} \left(\frac{3\mathrm{\pi }}{2}+ \mathrm{x}\right) \mathrm{cos}\left(2\mathrm{\pi } + \mathrm{x}\right) \left[\mathrm{cot} \left(\frac{3\mathrm{\pi }}{2} - \mathrm{x}\right) + \mathrm{cot} \left(2\mathrm{\pi }+\mathrm{x}\right)\right]=1$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{cos}\left(\frac{3\mathrm{\pi }}{2} +\mathrm{x}\right) \mathrm{cos} \left(2\mathrm{\pi } + \mathrm{x}\right) \left[\mathrm{cot}\left(\frac{3\mathrm{\pi }}{2}-\mathrm{x}\right)+ \mathrm{cot}\left(2\mathrm{\pi } + \mathrm{x}\right)\right]\\ =\mathrm{sin}\mathrm{x}.\mathrm{cos} \mathrm{x}\left[\mathrm{tan} \mathrm{x}+ \mathrm{cot}\mathrm{x}\right]\\ =\mathrm{sin}\mathrm{x}.\mathrm{cos}\mathrm{x}\left[\frac{\mathrm{sin} \mathrm{x}}{\mathrm{cos} \mathrm{x}} + \frac{\mathrm{cos} \mathrm{x}}{\mathrm{sin} \mathrm{x}}\right]\\ =\mathrm{sin}\mathrm{x}.\mathrm{cos}\mathrm{x}\left[\frac{{\mathrm{sin}}^{2}\mathrm{x}+ {\mathrm{cos}}^{2}\mathrm{x}}{\mathrm{sin}\mathrm{x}.\mathrm{cos} \mathrm{x}}\right]\end{array}$ $\begin{array}{l}={\mathrm{sin}}^{2}\mathrm{x} +{\mathrm{cos}}^{2}\mathrm{x}\\ = 1= \mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.19 sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos x

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}. : \text{sin}\left(\text{n\hspace{0.17em}}+\text{1}\right)\text{x sin\hspace{0.17em}}\left(\text{n}+ \text{2}\right) \text{x}\mathrm{}+\text{cos}\left(\text{n\hspace{0.17em}}+ \text{1}\right) \text{x cos\hspace{0.17em}}\left(\text{n\hspace{0.17em}}+\text{2}\right)\text{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{cos}\left(\text{n}+ \text{1}\right) \text{x cos\hspace{0.17em}}\left(\text{n\hspace{0.17em}}+ \text{2}\right) \text{x\hspace{0.17em}}+ \text{sin}\left(\text{n\hspace{0.17em}}+ \text{1}\right)\text{x sin\hspace{0.17em}}\left(\text{n}+\text{2}\right)\text{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{cos}\left\{\left(\mathrm{n}+ 1\right)\mathrm{x}- \left(\mathrm{n} + 2\right) \mathrm{x}\right\}\\ \left[\because \mathrm{cosA}\mathrm{cosB}-\mathrm{sinA}\mathrm{sinB} = \mathrm{cos}\left(\mathrm{A}+ \mathrm{B}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}= \text{cos}\left(\mathrm{nx}+\mathrm{x} -\mathrm{nx} - 2\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}= \text{cos\hspace{0.17em}}\left(-\mathrm{x}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \mathrm{cos}\left(-\mathrm{A}\right) = \mathrm{cosA}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}= \mathrm{cos}\mathrm{x} = \mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.20

$\mathrm{cos} \left(\frac{3\mathrm{\pi }}{4}+\mathrm{x}\right)-\mathrm{cos} \left(\frac{3\mathrm{\pi }}{4} -\mathrm{x}\right)= -\sqrt{2}\mathrm{sin}\mathrm{x}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \mathrm{cos} \left(\frac{3\mathrm{\pi }}{4} + \mathrm{x}\right) -\mathrm{cos} \left(\frac{3\mathrm{\pi }}{4}- \mathrm{x}\right)\\ = -2 \mathrm{sin} \left(\frac{\left(\frac{3\mathrm{\pi }}{4}+ \mathrm{x}\right) + \left(\frac{3\mathrm{\pi }}{4} - \mathrm{x}\right)}{2}\right)\mathrm{sin} \left(\frac{\left(\frac{3\mathrm{\pi }}{4}+ \mathrm{x}\right) - \left(\frac{3\mathrm{\pi }}{4}- \mathrm{x}\right)}{2}\right)\\ = -2\mathrm{sin}\left(\frac{2 \left(\frac{3\mathrm{\pi }}{4}\right)}{2}\right) \mathrm{sin}\left(\frac{2\mathrm{x}}{2}\right)\end{array}$ $\begin{array}{l}= -2 \mathrm{sin} \left(\mathrm{\pi } -\frac{\mathrm{\pi }}{4}\right) \mathrm{sin}\mathrm{x}\\ = -2\mathrm{sin}\left(\mathrm{\pi } -\frac{\mathrm{\pi }}{4}\right) \mathrm{sin} \mathrm{x}\\ = -2\mathrm{sin} \frac{\mathrm{\pi }}{4} × \mathrm{sin}\mathrm{x}\\ =-2 ×\frac{1}{\sqrt{2}} ×\mathrm{sin}\mathrm{x}\\ =-\sqrt{2} \mathrm{sin} \mathrm{x}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.21 Prove that:
sin2 6x – sin2 4x = sin 2x sin 10x

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}. = {\text{sin}}^{\text{2}}\text{\hspace{0.17em}6x\hspace{0.17em}}- {\text{sin\hspace{0.17em}}}^{\text{2}}\text{\hspace{0.17em}4x}\\ \text{\hspace{0.17em}\hspace{0.17em}}= \left(\mathrm{sin}6\mathrm{x} + \mathrm{sin} 4\mathrm{x}\right) \left(\mathrm{sin} 6\mathrm{x}- \mathrm{sin}4\mathrm{x}\right)\left[\because {\mathrm{a}}^{2} - {\mathrm{b}}^{2} = \left(\mathrm{a} - \mathrm{b}\right) \left(\mathrm{a} + \mathrm{b}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\left\{2 \mathrm{sin} \left(\frac{6\mathrm{x} +4\mathrm{x}}{2}\right)\mathrm{cos} \left(\frac{6\mathrm{x} - 4\mathrm{x}}{2}\right)\right\}\left\{2\mathrm{cos}\left(\frac{6\mathrm{x}+4\mathrm{x}}{2}\right)\mathrm{sin} \left(\frac{6\mathrm{x} -4\mathrm{x}}{2}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}}= \left\{2 \mathrm{sin} 5\mathrm{x}\mathrm{cos}\mathrm{x}\right\} \left\{2 \mathrm{cos} 5\mathrm{x} \mathrm{sin}\mathrm{x}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}}= \left(2\mathrm{sin} \mathrm{x}\mathrm{cos}\mathrm{x}\right)\left(2 \mathrm{sin} 5\mathrm{x} \mathrm{cos} 5\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{sin} 2\mathrm{x}.\mathrm{sin}10\mathrm{x}\left[\because \mathrm{sin}2\mathrm{x}=2\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{x}\right]\\ \text{\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{Hence}\text{proved.}\end{array}$

Q.22 Prove that:
cos2 2x – cos2 6x = sin 4x sin 8x

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}. ={\mathrm{cos}}^{\text{2}} \text{2x\hspace{0.17em}}- {\mathrm{cos}}^{\text{2}} \text{6x}\\ \text{\hspace{0.17em}\hspace{0.17em}}= \left(\mathrm{cos} 2\mathrm{x} + \mathrm{cos}6\mathrm{x}\right)\left(\mathrm{cos}2\mathrm{x} - \mathrm{cos}6\mathrm{x}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}= \left\{2\mathrm{cos} \left(\frac{2\mathrm{x} +6\mathrm{x}}{2}\right)\mathrm{cos}\left(\frac{2\mathrm{x} -6\mathrm{x}}{2}\right)\right\} \left\{-2 \mathrm{sin}\left(\frac{2\mathrm{x} -6\mathrm{x}}{2}\right) \mathrm{sin} \left(\frac{2\mathrm{x} +6\mathrm{x}}{2}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}}= \left\{2\mathrm{cos} 4\mathrm{x} \mathrm{cos}\left(-2\mathrm{x}\right)\right\} \left\{-2 \mathrm{sin} \left(-2\mathrm{x}\right) \mathrm{sin}4\mathrm{x}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}}= \left(2 \mathrm{sin} 2\mathrm{x} \mathrm{cos} 2\mathrm{x}\right)\left(2 \mathrm{sin} 4\mathrm{x} \mathrm{cos}4\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}= \mathrm{sin} 4\mathrm{x}.\mathrm{sin} 8\mathrm{x} \left[\because \mathrm{sin}2\mathrm{x} = 2\mathrm{sin} \mathrm{x}\mathrm{cos}\mathrm{x}\right]\\ \text{\hspace{0.17em}}= \mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{Hence}\text{\hspace{0.17em}proved.}\end{array}$

Q.23 Prove that:
sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x

Ans.

$\begin{array}{l}\text{sin\hspace{0.17em}2x\hspace{0.17em}}+\text{2 sin 4x}+\text{sin 6x\hspace{0.17em}}= \text{2 sin 4x\hspace{0.17em}}+ \text{sin 6x}+ \text{sin\hspace{0.17em}2x}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\text{2 sin\hspace{0.17em}4x\hspace{0.17em}}+ 2 \mathrm{sin}\left(\frac{6\mathrm{x} +2\mathrm{x}}{2}\right)\mathrm{cos} \left(\frac{6\mathrm{x} -2\mathrm{x}}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}= \text{2\hspace{0.17em}sin\hspace{0.17em}4x\hspace{0.17em}}+2\mathrm{sin} 4\mathrm{x} \mathrm{cos} 2\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\text{2\hspace{0.17em}sin 4x}\left(1+\mathrm{cos} 2\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\text{2\hspace{0.17em}sin 4x\hspace{0.17em}}\left(2 {\mathrm{cos}}^{2}\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}= 4{\mathrm{cos}}^{2}\mathrm{x}\text{\hspace{0.17em}sin 4x}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.24 Prove that:
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Ans.

$\begin{array}{l}\mathrm{We}\text{have, cot 4x\hspace{0.17em}}\left(\text{sin\hspace{0.17em}5x\hspace{0.17em}}+ \text{sin\hspace{0.17em}3x}\right)=\text{cot\hspace{0.17em}x\hspace{0.17em}}\left(\text{sin\hspace{0.17em}5x\hspace{0.17em}}–\text{sin\hspace{0.17em}3x}\right)\\ \mathrm{The}\text{given equation can be written as:}\\ \frac{\text{sin 5x}+\text{sin\hspace{0.17em}3x}}{\text{sin\hspace{0.17em}5x}–\text{sin 3x}}=\frac{\text{cot\hspace{0.17em}x}}{\text{cot 4x}}\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}:\frac{\text{sin\hspace{0.17em}5x\hspace{0.17em}}+ \text{sin\hspace{0.17em}3x}}{\text{sin 5x\hspace{0.17em}}– \text{sin 3x}}= \frac{2\mathrm{sin} \left(\frac{5\mathrm{x} +3\mathrm{x}}{2}\right) \mathrm{cos} \left(\frac{5\mathrm{x}-3\mathrm{x}}{2}\right)}{\text{2 cos\hspace{0.17em}}\frac{5\mathrm{x} + 3\mathrm{x}}{2}\text{sin}\frac{5\mathrm{x}- 3\mathrm{x}}{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}4\mathrm{x} \mathrm{cos}\mathrm{x}}{\mathrm{cos} 4\mathrm{x} \text{sin\hspace{0.17em}}\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\frac{\mathrm{cos} \mathrm{x}}{\mathrm{sin}\mathrm{x}}\right)}{\left(\frac{\mathrm{cos}4\mathrm{x}}{\mathrm{sin}4\mathrm{x}}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}}= \frac{\mathrm{cot}\mathrm{x}}{\mathrm{cot} 4\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{Hence}\text{proved.}\end{array}$

Q.25

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \frac{\mathrm{cos}9\mathrm{x}-\mathrm{cos}5\mathrm{x}}{\mathrm{sin}17\mathrm{x}-\mathrm{sin}3\mathrm{x}}=-\frac{\mathrm{sin}2\mathrm{x}}{\mathrm{cos}10\mathrm{x}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\mathrm{cos}9\mathrm{x}-\mathrm{cos}5\mathrm{x}}{\mathrm{sin}17\mathrm{x}-\mathrm{sin}3\mathrm{x}}=\frac{-2\mathrm{sin}\left(\frac{9\mathrm{x}+5\mathrm{x}}{2}\right)\mathrm{sin}\left(\frac{9\mathrm{x}-5\mathrm{x}}{2}\right)}{2\mathrm{cos}\left(\frac{17\mathrm{x}+3\mathrm{x}}{2}\right)\mathrm{sin}\left(\frac{17\mathrm{x}-3\mathrm{x}}{2}\right)}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\because \mathrm{cosA}-\mathrm{cosB}=-2\mathrm{sin}\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{sinA}-\mathrm{sinB}=-2\mathrm{cos}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\end{array}\right]\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=-\frac{\mathrm{sin}\left(7\mathrm{x}\right)\mathrm{sin}\left(2\mathrm{x}\right)}{\mathrm{cos}\left(10\mathrm{x}\right)\mathrm{sin}\left(7\mathrm{x}\right)}\\ =-\frac{\mathrm{sin}2\mathrm{x}}{\mathrm{cos}10\mathrm{x}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{​ proved.}\end{array}$

Q.26

$\begin{array}{l}\mathrm{Prove}\text{ }\mathrm{that}:\\ \frac{\mathrm{sin}5\text{x}+\mathrm{sin}3\text{x}}{\mathrm{cos}5\text{x}+\mathrm{cos}3\text{x}}=\mathrm{tan}4\text{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\mathrm{sin}5\mathrm{x}+\mathrm{sin}3\mathrm{x}}{\mathrm{cos}5\mathrm{x}+\mathrm{cos}3\mathrm{x}}=\frac{2\mathrm{sin}\left(\frac{5\mathrm{x}+3\mathrm{x}}{2}\right)\mathrm{cos}\left(\frac{5\mathrm{x}-3\mathrm{x}}{2}\right)}{2\mathrm{cos}\left(\frac{5\mathrm{x}+3\mathrm{x}}{2}\right)\mathrm{cos}\left(\frac{5\mathrm{x}-3\mathrm{x}}{2}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}4\mathrm{xcosx}}{\mathrm{cos}4\mathrm{xcosx}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{tan}4\mathrm{x}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.27

$\begin{array}{l}\mathrm{Prove}\text{ }\mathrm{that}:\\ \frac{\mathrm{sinx}-\mathrm{siny}}{\mathrm{cosx}+\mathrm{cosy}}=\mathrm{tan}\left(\frac{\text{x}-\text{y}}{2}\right)\end{array}$

Ans.

$\begin{array}{l}\text{L}.\text{H}.\text{S}.:\\ \frac{\mathrm{sinx}-\mathrm{siny}}{\mathrm{cosx}+\mathrm{cosy}}=\frac{2\mathrm{cos}\left(\frac{\text{x}+\text{y}}{2}\right)\mathrm{sin}\left(\frac{\text{x}-\text{y}}{2}\right)}{2\mathrm{cos}\left(\frac{\text{x}+\text{y}}{2}\right)\mathrm{cos}\left(\frac{\text{x}-\text{y}}{2}\right)}\\ \text{ }=\frac{\mathrm{sin}\left(\frac{\text{x}-\text{y}}{2}\right)}{\mathrm{cos}\left(\frac{\text{x}-\text{y}}{2}\right)}\\ \text{ }=\mathrm{tan}\left(\frac{\text{x}-\text{y}}{2}\right)\\ \text{ }=\text{R}.\text{H}.\text{S}.\text{ }\mathrm{Hence}\text{ proved.}\end{array}$

Q.28

$\begin{array}{l}\mathrm{Prove}\text{ }\mathrm{that}:\\ \frac{\mathrm{sinx}+\mathrm{sin}3\text{x}}{\mathrm{cosx}+\mathrm{cos}3\text{x}}=\mathrm{tan}\text{2x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\mathrm{sinx}+\mathrm{sin}3\mathrm{x}}{\mathrm{cosx}+\mathrm{cos}3\mathrm{x}}=\frac{\mathrm{sin}3\mathrm{x}+\mathrm{sinx}}{\mathrm{cos}3\mathrm{x}+\mathrm{cosx}}\\ =\frac{2\mathrm{sin}\left(\frac{3\mathrm{x}+\mathrm{x}}{2}\right)\mathrm{cos}\left(\frac{3\mathrm{x}-\mathrm{x}}{2}\right)}{2\mathrm{cos}\left(\frac{3\mathrm{x}+\mathrm{x}}{2}\right)\mathrm{cos}\left(\frac{3\mathrm{x}-\mathrm{x}}{2}\right)}\\ =\frac{\mathrm{sin}2\mathrm{x}}{\mathrm{cos}2\mathrm{x}}\\ =\mathrm{tan}2\mathrm{x}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{\hspace{0.17em}proved.}\end{array}$

Q.29

$\begin{array}{l}\mathbf{Prove}\mathbf{ }\mathbf{that}\mathbf{:}\\ \frac{\mathbf{sin}\mathbf{ }\mathbf{x}\mathbf{}\mathbf{–}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{3}\mathbf{x}}{{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\mathbf{ }\mathbf{–}\mathbf{}{\mathbf{cos}}^{\mathbf{2}}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\mathrm{sin} \mathrm{x} -\mathrm{sin}3\mathrm{x}}{{\mathrm{sin}}^{2}\mathrm{x} - {\mathrm{cos}}^{2}\mathrm{x}}=\frac{-\left(\mathrm{sin} 3\mathrm{x}-\mathrm{sin}\mathrm{x}\right)}{-\left({\mathrm{cos}}^{2}\mathrm{x} -{\mathrm{sin}}^{2}\mathrm{x}\right)}\\ =\frac{2 \mathrm{cos}\left(\frac{3\mathrm{x} +\mathrm{x}}{2}\right) \mathrm{sin}\left(\frac{3\mathrm{x}-\mathrm{x}}{2}\right)}{\mathrm{cos} 2\mathrm{x}}\\ = \frac{2 \mathrm{cos} 2\mathrm{x} \mathrm{sin}\mathrm{x}}{\mathrm{cos}2\mathrm{x}}\\ = 2 \mathrm{sin}\mathrm{x}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.30

$\begin{array}{l}\mathbf{Prove}\mathbf{ }\mathbf{that}\mathbf{:}\\ \frac{\mathbf{cos}\mathbf{}\mathbf{4}\mathbf{x}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{cos}\mathbf{ }\mathbf{3}\mathbf{x}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}}{\mathbf{sin}\mathbf{ }\mathbf{4}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{3}\mathbf{x}\mathbf{ }\mathbf{+}\mathbf{}\mathbf{sin}\mathbf{ }\mathbf{2}\mathbf{x}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{cot}\mathbf{}\mathbf{3}\mathbf{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\mathrm{cos} 4\mathrm{x} +\mathrm{cos}3\mathrm{x}+\mathrm{cos} 2\mathrm{x}}{\mathrm{sin}4\mathrm{x}+\mathrm{sin} 3\mathrm{x}+\mathrm{sin}2\mathrm{x}} =\frac{\mathrm{cos}4\mathrm{x}+\mathrm{cos}2\mathrm{x} +\mathrm{cos}3\mathrm{x}}{\mathrm{sin}4\mathrm{x} +\mathrm{sin}2\mathrm{x} +\mathrm{sin}3\mathrm{x}}\\ \text{\hspace{0.17em}}=\frac{2\mathrm{cos} \left(\frac{4\mathrm{x}+2\mathrm{x}}{2}\right)\mathrm{cos} \left(\frac{4\mathrm{x}-2\mathrm{x}}{2}\right)+\mathrm{cos}3\mathrm{x}}{2\mathrm{sin}\left(\frac{4\mathrm{x}+2\mathrm{x}}{2}\right)\mathrm{cos} \left(\frac{4\mathrm{x} -2\mathrm{x}}{2}\right) +\mathrm{sin}3\mathrm{x}}\\ \text{\hspace{0.17em}}=\frac{2 \mathrm{cos}3\mathrm{x}\mathrm{cos}\mathrm{x}+ \mathrm{cos} 3\mathrm{x}}{2\mathrm{sin}3\mathrm{x}\mathrm{cos}\mathrm{x} +\mathrm{sin}3\mathrm{x}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}=\frac{\mathrm{cos}3\mathrm{x}\left(2\mathrm{cos} \mathrm{x}+1\right)}{\mathrm{sin} 3\mathrm{x}\left(2\mathrm{cos}\mathrm{x} +1\right)}\\ \frac{\mathrm{cos}4\mathrm{x}+ \mathrm{cos} 3\mathrm{x}+\mathrm{cos}2\mathrm{x}}{\mathrm{sin}4\mathrm{x} +\mathrm{sin}3\mathrm{x} + \mathrm{sin} 2\mathrm{x}}=\mathrm{cot} 3\mathrm{x}= \mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{​ proved.}\end{array}$

Q.31 Prove that:
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Ans.

$\begin{array}{l}\because \mathrm{cot}3\mathrm{x}=\mathrm{cot}\left(2\mathrm{x}+\mathrm{x}\right)\\ =\frac{\mathrm{cot}2\mathrm{x}\mathrm{cot}\mathrm{x}-1}{\mathrm{cot}2\mathrm{x}+\mathrm{cot}\mathrm{x}}\\ \mathrm{cot}3\mathrm{x}\left(\mathrm{cot}2\mathrm{x}+\mathrm{cot}\mathrm{x}\right)=\mathrm{cot}2\mathrm{x}\mathrm{cot}\mathrm{x}-1\\ \mathrm{cot}3\mathrm{x}\mathrm{cot}2\mathrm{x}+\mathrm{cot}3\mathrm{x}\mathrm{cot}\mathrm{x}=\mathrm{cot}2\mathrm{x}\mathrm{cot}\mathrm{x}-1\\ 1=\mathrm{cot}\mathrm{x}\mathrm{cot}2\mathrm{x}-\mathrm{cot}2\mathrm{x}\mathrm{cot}3\mathrm{x}-\mathrm{cot}3\mathrm{x}\mathrm{cot}\mathrm{x}\\ \mathrm{or}\mathrm{cot}\mathrm{x}\mathrm{cot}2\mathrm{x}-\mathrm{cot}2\mathrm{x}\mathrm{cot}3\mathrm{x}-\mathrm{cot}3\mathrm{x}\mathrm{cot}\mathrm{x}=1.\mathrm{Hence}\text{​ proved.}\end{array}$

Q.32

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \mathrm{tan}4\mathrm{x}=\frac{4\mathrm{tan}\mathrm{x}\left(1–{\mathrm{tan}}^{2}\mathrm{x}\right)}{1–6{\mathrm{tan}}^{2}\mathrm{x}+{\mathrm{tan}}^{4}\mathrm{x}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}:\\ \mathrm{tan}4\mathrm{x}=\mathrm{tan}2\left(2\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2\mathrm{tan}2\mathrm{x}}{1-{\mathrm{tan}}^{2}2\mathrm{x}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2\left(\frac{2\mathrm{tan}\mathrm{x}}{1-{\mathrm{tan}}^{2}\mathrm{x}}\right)}{1-{\left(\frac{2\mathrm{tan}\mathrm{x}}{1-{\mathrm{tan}}^{2}\mathrm{x}}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\frac{4\mathrm{tan}\mathrm{x}}{1-{\mathrm{tan}}^{2}\mathrm{x}}\right)}{\left\{\frac{{\left(1-{\mathrm{tan}}^{2}\mathrm{x}\right)}^{2}-4{\mathrm{tan}}^{2}\mathrm{x}}{{\left(1-{\mathrm{tan}}^{2}\mathrm{x}\right)}^{2}}\right\}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4\mathrm{tan}\mathrm{x}}{1-{\mathrm{tan}}^{2}\mathrm{x}}×\frac{{\left(1-{\mathrm{tan}}^{2}\mathrm{x}\right)}^{2}}{1-2{\mathrm{tan}}^{2}\mathrm{x}+{\mathrm{tan}}^{4}\mathrm{x}-4{\mathrm{tan}}^{2}\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4\mathrm{tan}\mathrm{x}\left(1–{\mathrm{tan}}^{2}\mathrm{x}\right)}{1–6{\mathrm{tan}}^{2}\mathrm{x}+{\mathrm{tan}}^{4}\mathrm{x}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.33 Prove that:
cos 4x = 1 – 8sin2 x cos2 x

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\mathrm{cos}4\mathrm{x}=\mathrm{cos}2\left(2\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-2{\mathrm{sin}}^{2}\left(2\mathrm{x}\right)\left[\because \mathrm{cos}2\mathrm{x}=1-{\mathrm{sin}}^{2}\mathrm{x}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-2{\left(2\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{x}\right)}^{2}\left[\because \mathrm{sin}2\mathrm{x}=2\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{x}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-2\left(4{\mathrm{sin}}^{2}\mathrm{x}{\mathrm{cos}}^{2}\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-8{\mathrm{sin}}^{2}\mathrm{x}{\mathrm{cos}}^{2}\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.34 Prove that:
cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\text{cos 6x}\\ \text{\hspace{0.17em}}=\mathrm{cos}3\left(2\mathrm{x}\right)\\ \text{\hspace{0.17em}}=4{\mathrm{cos}}^{3}2\mathrm{x}-3\mathrm{cos}2\mathrm{x}\\ \text{\hspace{0.17em}}=4{\left(2{\mathrm{cos}}^{2}\mathrm{x}-1\right)}^{3}-3\left(2{\mathrm{cos}}^{2}\mathrm{x}-1\right)\\ \text{\hspace{0.17em}}=4\left(8{\mathrm{cos}}^{6}\mathrm{x}-12{\mathrm{cos}}^{4}\mathrm{x}+6{\mathrm{cos}}^{2}\mathrm{x}-1\right)-6{\mathrm{cos}}^{2}\mathrm{x}+3\\ \text{\hspace{0.17em}}=32{\mathrm{cos}}^{6}\mathrm{x}-48{\mathrm{cos}}^{4}\mathrm{x}+24{\mathrm{cos}}^{2}\mathrm{x}-4-6{\mathrm{cos}}^{2}\mathrm{x}+3\\ \text{\hspace{0.17em}}=32{\mathrm{cos}}^{6}\mathrm{x}-24{\mathrm{cos}}^{4}\mathrm{x}+18{\mathrm{cos}}^{2}\mathrm{x}-1\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.35

$\begin{array}{l}\mathrm{Find}\text{ }\mathrm{the}\text{ }\mathrm{principal}\text{ }\mathrm{and}\text{ }\mathrm{general}\text{ }\mathrm{solutions}\text{ }\mathrm{of}\text{ }\mathrm{the}\text{ }\mathrm{following}\\ \mathrm{equations}:\mathrm{tan}\text{ }\mathrm{x}=\sqrt{3}\end{array}$

Ans.

$\begin{array}{l}\begin{array}{l}\mathrm{Since},\text{tan}\frac{\text{π}}{3}=\sqrt{3}\text{and tan}\frac{4\text{π}}{3}=\text{tan}\left(\text{π}+\frac{\text{π}}{3}\right)=\mathrm{tan}\frac{\text{π}}{3}=\sqrt{3}\\ \mathrm{Therefore},\text{principal solutions are x}=\frac{\text{π}}{3}\text{and}\frac{4\text{π}}{3}.\\ \text{General solution of tanx}=\sqrt{3}\\ \mathrm{tanx}=\sqrt{3}\\ \text{ }=\mathrm{tan}\frac{\text{π}}{3}\end{array}\\ \begin{array}{l}\mathrm{tanx}=\mathrm{tan}\left(\mathrm{n\pi }+\frac{\text{π}}{3}\right)\\ ⇒\text{ x}=\mathrm{n\pi }+\frac{\text{π}}{3},\mathrm{where}\text{ n}\in \text{Z}.\end{array}\end{array}$

Q.36

$\begin{array}{l}\mathrm{Find}\text{ }\mathrm{the}\text{ }\mathrm{principal}\text{ }\mathrm{and}\text{ }\mathrm{general}\text{ }\mathrm{solutions}\text{ }\mathrm{of}\text{ }\mathrm{the}\text{ }\mathrm{following}\\ \mathrm{equation}:\mathrm{sec}\text{ }\mathrm{x}=2\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{secx}=\text{2}⇒\mathrm{sec}\frac{\mathrm{\pi }}{3}=2\text{and sec}\frac{\text{5}\mathrm{\pi }}{3}=\mathrm{sec}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)=\mathrm{sec}\frac{\mathrm{\pi }}{3}=2\\ \mathrm{Therefore},\text{principal solutions are x}=\frac{\mathrm{\pi }}{3}\text{and}\frac{\text{5}\mathrm{\pi }}{3}.\\ \mathrm{G}\text{eneral solution:}\\ \mathrm{secx}=2\\ =\mathrm{sec}\frac{\mathrm{\pi }}{3}\\ =\mathrm{sec}\left(2\mathrm{n\pi }±\frac{\mathrm{\pi }}{3}\right)\\ \therefore \mathrm{x}=2\mathrm{n\pi }±\frac{\mathrm{\pi }}{3},\mathrm{where}\text{​ n}\in \mathrm{Z}.\end{array}$

Q.37

$\begin{array}{l}\mathrm{Find}\text{ }\mathrm{the}\text{ }\mathrm{principal}\text{ }\mathrm{and}\text{ }\mathrm{general}\text{ }\mathrm{solutions}\text{ }\mathrm{of}\text{ }\mathrm{the}\text{ }\mathrm{following}\\ \mathrm{equation}:\mathrm{cot}\text{ }\mathrm{x}=-\sqrt{3}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{cot\hspace{0.17em}x}=-\sqrt{3}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}cot\hspace{0.17em}x}=-\mathrm{cot}\frac{\mathrm{\pi }}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cot}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cot}\left(\frac{5\mathrm{\pi }}{6}\right)\\ \mathrm{and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}cot\hspace{0.17em}x}=\mathrm{cot}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cot}\left(\frac{11\mathrm{\pi }}{6}\right)\\ \mathrm{Therefore},\text{the principal solution are x}=\frac{5\mathrm{\pi }}{6}\text{and}\frac{11\mathrm{\pi }}{6}.\\ \mathrm{General}\text{solution:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cotx}=-\sqrt{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cot}\left(\frac{5\mathrm{\pi }}{6}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cot}\left(\mathrm{n\pi }+\frac{5\mathrm{\pi }}{6}\right)\\ ⇒\mathrm{x}=\mathrm{n\pi }+\frac{5\mathrm{\pi }}{6},\mathrm{where}\text{​ n}\in \mathrm{Z}.\end{array}$

Q.38

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{principal}\mathrm{and}\mathrm{general}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\mathrm{cosec}\mathrm{x}=-2\end{array}$

Ans.

$\mathrm{Since},\mathrm{cosec}\mathrm{x}=\mathrm{cosec}\left(\frac{\mathrm{\pi }}{6}\right)$ $\begin{array}{l}\therefore \mathrm{cosec}\mathrm{x}=-2\\ =-\mathrm{cosec}\left(\frac{\mathrm{\pi }}{6}\right)\\ =\mathrm{cosec}\left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\\ =\mathrm{cosec}\left(\frac{7\mathrm{\pi }}{6}\right)\\ \mathrm{and}\\ \mathrm{cosec}\mathrm{x}=\mathrm{cosec}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)\\ =\mathrm{cosec}\frac{11\mathrm{\pi }}{6}\\ \mathrm{Therefore},\text{the principal solutions are}\frac{7\mathrm{\pi }}{6},\frac{11\mathrm{\pi }}{6}.\\ \mathrm{General}\text{solution:}\\ \text{cosec x}=\mathrm{cosec}\left(\frac{7\mathrm{\pi }}{6}\right)\\ =\mathrm{cosec}\left(\mathrm{n\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{6}\right)\\ ⇒\mathrm{x}=\mathrm{n\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{6},\mathrm{n}\in \mathrm{Z}.\end{array}$

Q.39

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{for}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\mathrm{cos}4\mathrm{x}=\mathrm{cos}2\mathrm{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{​​\hspace{0.17em} have,}\\ \text{cos 4x}=\text{cos 2x}\end{array}$ $\begin{array}{l}=\mathrm{cos}\left(2\mathrm{n\pi }±2\mathrm{x}\right)\left[\begin{array}{l}\mathrm{The}\text{general solution of cos}\mathrm{\theta }=\mathrm{cos}\mathrm{\alpha }\\ ⇒\mathrm{\theta }=2\mathrm{n\pi }±\mathrm{\alpha }\end{array}\right]\\ ⇒4\mathrm{x}=2\mathrm{n\pi }±2\mathrm{x}\\ \mathrm{For}\text{\hspace{0.17em}}\left(+\right)\mathrm{ve}\text{sign:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\mathrm{x}=2\mathrm{n\pi }+2\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}=2\mathrm{n\pi }⇒\mathrm{x}=\mathrm{n\pi }\\ \mathrm{For}\left(-\right)\mathrm{ve}\text{sign:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\mathrm{x}=2\mathrm{n\pi }-2\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6\mathrm{x}=2\mathrm{n\pi }⇒\mathrm{x}=\frac{1}{3}\mathrm{n\pi }\\ \mathrm{The}\text{general solutions are x}=\frac{1}{3}\mathrm{n\pi }\text{and}\mathrm{n\pi },\text{where n}\in \mathrm{Z}.\end{array}$

Q.40

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{for}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\mathrm{cos}3\mathrm{x}+\mathrm{cos}\mathrm{x}-\mathrm{cos}2\mathrm{x}=0\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,\hspace{0.17em}}\mathrm{cos}\text{}3\mathrm{x}+\mathrm{cos}\mathrm{x}-\mathrm{cos}2\mathrm{x}=0\\ 2\mathrm{cos}\left(\frac{3\mathrm{x}+\mathrm{x}}{2}\right)\mathrm{cos}\left(\frac{3\mathrm{x}-\mathrm{x}}{2}\right)-\mathrm{cos}2\mathrm{x}=0\\ 2\mathrm{cos}2\mathrm{x}\mathrm{cos}\mathrm{x}-\mathrm{cos}2\mathrm{x}=0\\ \mathrm{cos}2\mathrm{x}\left(2\mathrm{cos}\mathrm{x}-1\right)=0\\ ⇒\mathrm{cos}2\mathrm{x}=0\text{​}⇒2\mathrm{x}=\left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{Z}\end{array}$ $\begin{array}{l}\text{or 2 cos x}-1=0⇒\mathrm{cos}\mathrm{x}=\frac{1}{2}=\mathrm{cos}\frac{\mathrm{\pi }}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}=2\mathrm{n\pi }±\frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z}.\end{array}$

Q.41

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{for}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\mathrm{sin}2\mathrm{x}+\mathrm{cos}\mathrm{x}=0\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\mathrm{sin}2\mathrm{x}+\mathrm{cos}\mathrm{x}=0\\ ⇒2\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{x}+\mathrm{cos}\mathrm{x}=0\\ ⇒\mathrm{cos}\mathrm{x}\left(2\mathrm{sin}\mathrm{x}+1\right)=0\\ \mathrm{Either}\mathrm{cos}\mathrm{x}=0⇒\mathrm{x}=\left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z}.\\ \mathrm{or}\left(2\mathrm{sin}\mathrm{x}+1\right)=0⇒\mathrm{sin}\mathrm{x}=-\frac{1}{2}\\ ⇒\mathrm{sin}\mathrm{x}=\mathrm{sin}\frac{7\mathrm{\pi }}{2}\\ ⇒\mathrm{x}=\mathrm{n\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z}.\\ \mathrm{Thus},\text{the general solution of given equation is:}\\ \mathrm{n\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{2}\text{​\hspace{0.17em}\hspace{0.17em}or}\left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z}.\end{array}$

Q.42

$\begin{array}{l}\mathrm{Find}\text{ }\mathrm{the}\text{ }\mathrm{general}\text{ }\mathrm{solution}\text{ }\mathrm{for}\text{ }\mathrm{each}\text{ }\mathrm{of}\text{ }\mathrm{the}\text{ }\mathrm{following}\\ \mathrm{equation}:\text{ }{\mathrm{sec}}^{2}\text{ }2\text{x}=1-\mathrm{tan}2\text{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},{\mathrm{sec}}^{2}2\mathrm{x}=1-\mathrm{tan}2\mathrm{x}\\ ⇒1+{\mathrm{tan}}^{2}2\mathrm{x}=1-\mathrm{tan}2\mathrm{x}\end{array}$ $\begin{array}{l}⇒{\mathrm{tan}}^{2}2\mathrm{x}+\mathrm{tan}2\mathrm{x}=0\\ ⇒\mathrm{tan}2\mathrm{x}\left(\mathrm{tan}2\mathrm{x}+1\right)=0\\ \mathrm{Either}\text{tan 2x}=\text{0}⇒2\mathrm{x}=\mathrm{n\pi }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}=\frac{\mathrm{n\pi }}{2},\text{\hspace{0.17em}where n\hspace{0.17em}}\in \mathrm{Z}.\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan}\text{}2\mathrm{x}+1=0⇒\mathrm{tan}2\mathrm{x}=-1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{tan}2\mathrm{x}=-\mathrm{tan}\frac{\mathrm{\pi }}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{tan}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{tan}2\mathrm{x}=\mathrm{tan}\frac{3\mathrm{\pi }}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒2\mathrm{x}=\mathrm{n\pi }+\frac{3\mathrm{\pi }}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}=\frac{\mathrm{n\pi }}{2}+\frac{3\mathrm{\pi }}{8},\text{where n}\in \mathrm{Z}.\\ \mathrm{The}\text{general solution of the given equation is:}\\ \mathrm{x}=\frac{\mathrm{n\pi }}{2},\frac{\mathrm{n\pi }}{2}+\frac{3\mathrm{\pi }}{8},\text{\hspace{0.17em}where n\hspace{0.17em}}\in \mathrm{Z}.\end{array}$

Q.43

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{for}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\mathrm{sin}\mathrm{x}+\mathrm{sin}3\mathrm{x}+\mathrm{sin}5\mathrm{x}=0\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{sin}\mathrm{x}+\mathrm{sin}3\mathrm{x}+\mathrm{sin}5\mathrm{x}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{sin}\text{}5\mathrm{x}+\mathrm{sin}\mathrm{x}+\mathrm{sin}3\mathrm{x}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{sin}\left(\frac{5\mathrm{x}+\mathrm{x}}{2}\right)\mathrm{cos}\left(\frac{5\mathrm{x}-\mathrm{x}}{2}\right)+\mathrm{sin}3\mathrm{x}=0\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{sin}3\mathrm{x}\mathrm{cos}2\mathrm{x}+\mathrm{sin}3\mathrm{x}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{sin}\text{}3\mathrm{x}\left(2\mathrm{cos}2\mathrm{x}+1\right)=0\\ \mathrm{Either}\text{sin3x}=\text{0}⇒3\mathrm{x}=\mathrm{n\pi }\\ ⇒\mathrm{x}=\frac{\mathrm{n\pi }}{3},\text{\hspace{0.17em}}\mathrm{n}\in \mathrm{Z}.\\ \text{or 2 cos 2x}+1=0⇒\mathrm{cos}2\mathrm{x}=-\frac{1}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}=2\mathrm{n\pi }±\frac{2\mathrm{\pi }}{3}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{n\pi }±\frac{\mathrm{\pi }}{3},\text{\hspace{0.17em}}\mathrm{n}\in \mathrm{Z}.\\ \mathrm{Therefore},\text{​​ the general solution of the given equations is:}\\ \mathrm{x}=\frac{\mathrm{n\pi }}{3}\text{​​}\mathrm{or}\text{\hspace{0.17em}}\mathrm{n\pi }±\frac{\mathrm{\pi }}{3},\text{\hspace{0.17em}}\mathrm{n}\in \mathrm{Z}.\text{\hspace{0.17em}}\end{array}$

Q.44 In any triangle ABC, if a = 18, b = 24, c = 30, find

1. cos A, cos B, cos C
2. sin A, sin B, sin C

Ans.

$\begin{array}{l}1.\\ \mathrm{Since},\text{ }\mathrm{cos}\text{ }\mathrm{A}=\frac{{\text{b}}^{2}+{\text{c}}^{2}-{\text{a}}^{2}}{2\mathrm{bc}}\\ \text{ }=\frac{{\left(24\right)}^{2}+{\left(30\right)}^{2}-{\left(18\right)}^{2}}{2\left(24\right)\left(30\right)}\\ \text{ }=\frac{576+900-324}{1440}\\ \text{ }=\frac{1152}{1440}\\ \text{ }=\frac{4}{5}\\ \mathrm{cos}\text{ }\mathrm{B}=\frac{{\text{c}}^{2}+{\text{a}}^{2}-{\text{b}}^{2}}{2\mathrm{ca}}\\ \text{ }=\frac{{\left(30\right)}^{2}+{\left(18\right)}^{2}-{\left(24\right)}^{2}}{2\left(30\right)\left(18\right)}\\ \text{ }=\frac{900+324-576}{1080}\\ \text{ }=\frac{648}{1080}\\ \text{ }=\frac{3}{5}\\ \mathrm{cos}\text{ }\mathrm{C}=\frac{{\text{a}}^{2}+{\text{b}}^{2}-{\text{c}}^{2}}{2\mathrm{ab}}\\ \text{ }=\frac{{\left(18\right)}^{2}+{\left(24\right)}^{2}-{\left(30\right)}^{2}}{2\left(18\right)\left(24\right)}\\ \text{ }=\frac{324+576-900}{864}\\ \text{ }=\frac{0}{864}\\ \text{ }=0\end{array}$

$\begin{array}{l}2.\\ \mathrm{Since},\text{ }\mathrm{cos}\text{ }\mathrm{A}=\frac{{\text{b}}^{2}+{\text{c}}^{2}-{\text{a}}^{2}}{2\mathrm{bc}}\\ \text{ }=\frac{{\left(24\right)}^{2}+{\left(30\right)}^{2}-{\left(18\right)}^{2}}{2\left(24\right)\left(30\right)}\\ \text{ }=\frac{576+900-324}{1440}\\ \text{ }=\frac{1152}{1440}\\ \text{ }=\frac{4}{5}\\ \mathrm{and}\text{ }\mathrm{sinA}=\sqrt{1-{\mathrm{cos}}^{2}\text{A}}\\ \text{ }=\sqrt{1-{\left(\frac{4}{5}\right)}^{2}}\\ \text{ }=\sqrt{1-\frac{16}{25}}\\ \text{ }=\frac{3}{5}\\ \mathrm{Since},\text{ }\frac{\mathrm{sin}\text{ }\mathrm{A}}{\text{a}}=\frac{\mathrm{sin}\text{ }\mathrm{B}}{\text{b}}=\frac{\mathrm{sin}\text{ }\mathrm{C}}{\text{c}}\\ ⇒\text{ }\frac{\left(\frac{3}{5}\right)}{18}=\frac{\mathrm{sin}\text{ }\mathrm{B}}{24}=\frac{\mathrm{sin}\text{ }\mathrm{C}}{30}\\ ⇒\text{ }\frac{\left(\frac{3}{5}\right)}{18}=\frac{\mathrm{sin}\text{ }\mathrm{B}}{24}\text{and}\frac{\left(\frac{3}{5}\right)}{18}=\frac{\mathrm{sin}\text{ }\mathrm{C}}{30}\\ ⇒\text{ }\frac{3×24}{5×18}=\mathrm{sin}\text{ }\mathrm{B}\text{ and }\frac{3×30}{5×18}=\mathrm{sin}\text{ }\mathrm{C}\\ ⇒\text{ }\frac{4}{5}=\mathrm{sinB}\text{and }1=\mathrm{sin}\text{ }\mathrm{C}\\ \mathrm{Thus},\text{}\mathrm{sinA}=\frac{3}{5},\text{}\mathrm{sinB}=\frac{4}{5},\text{}\mathrm{sinC}=1.\end{array}$

Q.45

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{c}}\\ =\frac{\mathrm{k}\mathrm{sin}\mathrm{A}-\mathrm{k}\mathrm{sin}\mathrm{B}}{\mathrm{k}\mathrm{sin}\mathrm{C}}\left[\because \frac{\mathrm{a}}{\mathrm{sin}\mathrm{A}}=\frac{\mathrm{b}}{\mathrm{sin}\mathrm{B}}=\frac{\mathrm{c}}{\mathrm{sin}\mathrm{C}}=\mathrm{k}\left(\mathrm{let}\right)\right]\\ =\frac{\mathrm{sin}\mathrm{A}-\mathrm{sin}\mathrm{B}}{\mathrm{sin}\mathrm{C}}\\ =\frac{2\mathrm{cos}\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{2\mathrm{sin}\frac{\mathrm{C}}{2}\mathrm{cos}\frac{\mathrm{C}}{2}}\\ =\frac{\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{C}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{\mathrm{sin}\frac{\mathrm{C}}{2}\mathrm{cos}\frac{\mathrm{C}}{2}}\\ =\frac{\mathrm{sin}\left(\frac{\mathrm{C}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{\mathrm{sin}\frac{\mathrm{C}}{2}\mathrm{cos}\frac{\mathrm{C}}{2}}\end{array}$ $\begin{array}{l}=\frac{\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{\mathrm{cos}\frac{\mathrm{C}}{2}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{it is proved.}\end{array}$

Q.46

$\begin{array}{l}\text{For any triangle ABC},\text{prove that}\\ \frac{\mathrm{a}-\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{\mathrm{cos}\frac{\mathrm{C}}{2}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{c}}\\ =\frac{\mathrm{k}\mathrm{sin}\mathrm{A}-\mathrm{k}\mathrm{sin}\mathrm{B}}{\mathrm{k}\mathrm{sin}\mathrm{C}}\left[\because \frac{\mathrm{a}}{\mathrm{sin}\mathrm{A}}=\frac{\mathrm{b}}{\mathrm{sin}\mathrm{B}}=\frac{\mathrm{c}}{\mathrm{sin}\mathrm{C}}=\mathrm{k}\left(\mathrm{let}\right)\right]\\ =\frac{\mathrm{sin}\mathrm{A}-\mathrm{sin}\mathrm{B}}{\mathrm{sin}\mathrm{C}}\\ =\frac{2\mathrm{cos}\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{2\mathrm{sin}\frac{\mathrm{C}}{2}\mathrm{cos}\frac{\mathrm{C}}{2}}\\ =\frac{\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{C}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{\mathrm{sin}\frac{\mathrm{C}}{2}\mathrm{cos}\frac{\mathrm{C}}{2}}\\ =\frac{\mathrm{sin}\left(\frac{\mathrm{C}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{\mathrm{sin}\frac{\mathrm{C}}{2}\mathrm{cos}\frac{\mathrm{C}}{2}}\end{array}$ $\begin{array}{l}=\frac{\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{\mathrm{cos}\frac{\mathrm{C}}{2}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{it is proved.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\\ =\frac{\mathrm{k}\mathrm{sin}\mathrm{A}+\mathrm{k}\mathrm{sin}\mathrm{B}}{\mathrm{k}\mathrm{sin}\mathrm{C}}\left[\because \frac{\mathrm{a}}{\mathrm{sin}\mathrm{A}}=\frac{\mathrm{b}}{\mathrm{sin}\mathrm{B}}=\frac{\mathrm{c}}{\mathrm{sin}\mathrm{C}}=\mathrm{k}\left(\mathrm{let}\right)\right]\\ =\frac{\mathrm{sin}\mathrm{A}+\mathrm{sin}\mathrm{B}}{\mathrm{sin}\mathrm{C}}\\ =\frac{2\mathrm{sin}\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\mathrm{cos}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{2\mathrm{sin}\frac{\mathrm{C}}{2}\mathrm{cos}\frac{\mathrm{C}}{2}}\\ =\frac{2\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{C}}{2}\right)\mathrm{cos}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{2\mathrm{sin}\frac{\mathrm{C}}{2}\mathrm{cos}\frac{\mathrm{C}}{2}}\\ =\frac{\mathrm{cos}\left(\frac{\mathrm{C}}{2}\right)\mathrm{cos}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{\mathrm{sin}\frac{\mathrm{C}}{2}\mathrm{cos}\frac{\mathrm{C}}{2}}\end{array}$ $\begin{array}{l}=\frac{\mathrm{cos}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{\mathrm{sin}\frac{\mathrm{C}}{2}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{it is proved.}\end{array}$

Q.47

$\begin{array}{l}\text{For any triangle ABC},\text{prove that}\\ \mathrm{sin}\frac{\mathrm{B}-\mathrm{C}}{2}=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{a}}\mathrm{cos}\frac{\mathrm{A}}{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{b}-\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{k}\mathrm{sin}\mathrm{B}-\mathrm{k}\mathrm{sin}\mathrm{C}}{\mathrm{k}\mathrm{sin}\mathrm{A}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \frac{\mathrm{a}}{\mathrm{sin}\mathrm{A}}=\frac{\mathrm{b}}{\mathrm{sin}\mathrm{B}}=\frac{\mathrm{c}}{\mathrm{sin}\mathrm{C}}=\mathrm{k}\left(\mathrm{let}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}\mathrm{B}-\mathrm{sin}\mathrm{A}}{\mathrm{sin}\mathrm{A}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2\mathrm{cos}\left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{2\mathrm{sin}\frac{\mathrm{A}}{2}\mathrm{cos}\frac{\mathrm{A}}{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{A}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{\mathrm{sin}\frac{\mathrm{A}}{2}\mathrm{cos}\frac{\mathrm{A}}{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}\left(\frac{\mathrm{A}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{\mathrm{sin}\frac{\mathrm{A}}{2}\mathrm{cos}\frac{\mathrm{A}}{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{\mathrm{cos}\frac{\mathrm{A}}{2}}\end{array}$ $\begin{array}{l}\mathrm{sin}\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{a}}\mathrm{cos}\frac{\mathrm{A}}{2}\\ \mathrm{Therefore},\text{it is proved.}\end{array}$

Q.48 For any triangle ABC, prove that a (b cos C – c cos B) = b2 – c2

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \mathrm{b}\text{ }\mathrm{cos}\text{ }\mathrm{C}-\mathrm{c}\text{ }\mathrm{cos}\text{ }\mathrm{B}=\text{b}×\frac{{\text{a}}^{2}+{\text{b}}^{2}-{\text{c}}^{2}}{2\mathrm{ab}}-\text{c}×\frac{{\text{a}}^{2}+{\text{c}}^{2}-{\text{b}}^{2}}{2\mathrm{ac}}\\ \text{ }=\frac{{\text{a}}^{2}+{\text{b}}^{2}-{\text{c}}^{2}}{2\text{a}}-\frac{{\text{a}}^{2}+{\text{c}}^{2}-{\text{b}}^{2}}{2\text{a}}\\ \text{ }=\frac{{\text{a}}^{2}+{\text{b}}^{2}-{\text{c}}^{2}-{\text{a}}^{2}-{\text{c}}^{2}+{\text{b}}^{2}}{2\text{a}}\\ \text{ }=\frac{2{\text{b}}^{2}-2{\text{c}}^{2}}{2\text{a}}\\ \text{a}\left(\mathrm{b}\text{ }\mathrm{cos}\text{ }\mathrm{C}-\mathrm{c}\text{ }\mathrm{cos}\text{ }\mathrm{B}\right)={\text{b}}^{2}-{\text{c}}^{2}\\ \mathrm{Hence},\text{it is proved.}\end{array}$

Q.49

$\begin{array}{l}\text{For any triangle ABC},\text{prove that}\\ \mathrm{a}\left(\mathrm{cos}\text{\hspace{0.17em}}\mathrm{C}-\mathrm{cos}\text{\hspace{0.17em}}\mathrm{B}\right)=2\left(\mathrm{b}-\mathrm{c}\right){\mathrm{cos}}^{2}\frac{\mathrm{A}}{2}\end{array}$

Ans.

$\begin{array}{l}\text{We }\mathrm{are}\text{ }\mathrm{given}\text{ }\mathrm{that}:\\ \text{ }\mathrm{a}\left(\mathrm{cos}\text{ }\mathrm{C}-\mathrm{cos}\text{ }\mathrm{B}\right)=2\left(\mathrm{b}-\mathrm{c}\right){\mathrm{cos}}^{2}\frac{\mathrm{A}}{2}\\ ⇒\text{ }\frac{\left(\mathrm{cos}\text{ }\mathrm{C}-\mathrm{cos}\text{ }\mathrm{B}\right)}{2{\mathrm{cos}}^{2}\frac{\mathrm{A}}{2}}=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{a}}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\left(\mathrm{cos}\text{ }\mathrm{C}-\mathrm{cos}\text{ }\mathrm{B}\right)}{2{\mathrm{cos}}^{2}\frac{\mathrm{A}}{2}}\\ \text{ }=\frac{\mathrm{cos}\frac{\text{A}}{2}\mathrm{sin}\left(\frac{\text{B}-\text{C}}{2}\right)}{{\mathrm{cos}}^{2}\frac{\text{A}}{2}}\left[\because \mathrm{sin}\left(\frac{\text{π}}{2}-\text{θ}\right)=\mathrm{sin\theta }\right]\\ \text{ }=\frac{\mathrm{cos}\left(\frac{\text{B}+\text{C}}{2}\right)}{\mathrm{cos}\left(\frac{\text{B}+\text{C}}{2}\right)}×\frac{\mathrm{sin}\left(\frac{\text{B}-\text{C}}{2}\right)}{\mathrm{cos}\frac{\text{A}}{2}}\text{ }\left[\mathrm{Multiply}\text{and divide by }\mathrm{cos}\left(\frac{\text{B}+\text{C}}{2}\right)\right]\\ \text{ }=\frac{2\mathrm{cos}\left(\frac{\text{B}+\text{C}}{2}\right)}{2\mathrm{cos}\left(\frac{\text{π}}{2}-\frac{\text{A}}{2}\right)}×\frac{\mathrm{sin}\left(\frac{\text{B}-\text{C}}{2}\right)}{\mathrm{cos}\frac{\text{A}}{2}}\\ \text{ }=\frac{2\mathrm{cos}\left(\frac{\text{B}+\text{C}}{2}\right)\mathrm{sin}\left(\frac{\text{B}-\text{C}}{2}\right)}{2\mathrm{sin}\left(\frac{\text{A}}{2}\right)\mathrm{cos}\frac{\text{A}}{2}}\\ \text{ }=\frac{\mathrm{sinB}-\mathrm{sinC}}{\mathrm{sinA}}\left[\begin{array}{l}\because \text{ }2\mathrm{cos}\left(\frac{\text{x}+\text{y}}{2}\right)\mathrm{sin}\left(\frac{\text{x}-\text{y}}{2}\right)=\mathrm{sin}\text{ }\mathrm{x}-\mathrm{sin}\text{ }\mathrm{y}\\ \mathrm{sin\theta }=2\mathrm{sin}\frac{\text{θ}}{2}\text{ }\mathrm{cos}\frac{\text{θ}}{2}\end{array}\right]\\ \text{ }=\frac{\mathrm{kb}-\mathrm{kc}}{\mathrm{ka}}\left[\frac{\mathrm{sinA}}{\text{a}}=\frac{\mathrm{sinB}}{\text{b}}=\frac{\mathrm{sinC}}{\text{c}}=\text{k }\left(\mathrm{Let}\right)\right]\\ \text{ }=\frac{\text{b}-\text{c}}{\text{a}}=\text{R}.\text{H}.\text{S}.\\ \mathrm{Thus},\text{a}\left(\mathrm{cos}\text{ C}-\mathrm{cos}\text{ B}\right)=2\left(\text{b}-\text{c}\right){\mathrm{cos}}^{2}\frac{\text{A}}{2}\text{is proved.}\end{array}$

Q.50

$\begin{array}{l}\text{For any triangle ABC},\text{prove that}\\ \frac{\mathrm{sin}\left(\mathrm{B}-\mathrm{C}\right)}{\mathrm{sin}\left(\mathrm{B}+\mathrm{C}\right)}=\frac{{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{k}}^{2}{\mathrm{sin}}^{2}\mathrm{B}-{\mathrm{k}}^{2}{\mathrm{sin}}^{2}\mathrm{C}}{{\mathrm{k}}^{2}{\mathrm{sin}}^{2}\mathrm{A}}\left[\because \frac{\mathrm{a}}{\mathrm{sinA}}=\frac{\mathrm{b}}{\mathrm{sinB}}=\frac{\mathrm{c}}{\mathrm{sinC}}=\mathrm{k}\left(\mathrm{let}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{sin}}^{2}\mathrm{B}-{\mathrm{sin}}^{2}\mathrm{C}}{{\mathrm{sin}}^{2}\mathrm{A}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}\left(\mathrm{B}+\mathrm{C}\right)\mathrm{sin}\left(\mathrm{B}-\mathrm{C}\right)}{{\mathrm{sin}}^{2}\mathrm{A}}\left[\because {\mathrm{sin}}^{2}\mathrm{x}-{\mathrm{sin}}^{2}\mathrm{y}=\mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{sin}\left(\mathrm{x}-\mathrm{y}\right)\right]\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}\left(\mathrm{\pi }-\mathrm{A}\right)\mathrm{sin}\left(\mathrm{B}-\mathrm{C}\right)}{{\mathrm{sin}}^{2}\mathrm{A}}\left[\because \mathrm{B}+\mathrm{C}=\mathrm{\pi }-\mathrm{A}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}\mathrm{A}\mathrm{sin}\left(\mathrm{B}-\mathrm{C}\right)}{{\mathrm{sin}}^{2}\mathrm{A}}\left[\because \mathrm{sin}\left(\mathrm{\pi }-\mathrm{A}\right)=\mathrm{sinA}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}\left(\mathrm{B}-\mathrm{C}\right)}{\mathrm{sin}\left\{\mathrm{\pi }-\left(\mathrm{B}+\mathrm{C}\right)\right\}}\left[\because \mathrm{A}=\mathrm{\pi }-\left(\mathrm{B}+\mathrm{C}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}\left(\mathrm{B}-\mathrm{C}\right)}{\mathrm{sin}\left(\mathrm{B}+\mathrm{C}\right)}\left[\because \mathrm{sin}\left(\mathrm{\pi }-\mathrm{\theta }\right)=\mathrm{sin\theta }\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{it is proved.}\end{array}$

Q.51

$\begin{array}{l}\text{For any triangle ABC},\text{prove that}\\ \left(\mathrm{b}+\mathrm{c}\right)\mathrm{cos}\frac{\mathrm{B}+\mathrm{C}}{2}=\mathrm{a}\mathrm{cos}\frac{\mathrm{B}-\mathrm{C}}{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{are given :}\left(\mathrm{b}+\mathrm{c}\right)\mathrm{cos}\text{}\frac{\mathrm{B}+\mathrm{C}}{2}\text{}=\mathrm{a}\mathrm{cos}\frac{\mathrm{B}-\mathrm{C}}{2}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\left(\mathrm{b}+\mathrm{c}\right)}{\mathrm{a}}\text{}=\frac{\mathrm{cos}\frac{\mathrm{B}-\mathrm{C}}{2}}{\mathrm{cos}\frac{\mathrm{B}+\mathrm{C}}{2}}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\text{\hspace{0.17em}\hspace{0.17em}}\frac{\left(\mathrm{b}+\mathrm{c}\right)}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}\mathrm{sin}\mathrm{B}+\mathrm{k}\mathrm{sin}\mathrm{C}}{\mathrm{k}\mathrm{sin}\mathrm{A}}\left[\because \frac{\mathrm{a}}{\mathrm{sin}\mathrm{A}}=\frac{\mathrm{b}}{\mathrm{sin}\mathrm{B}}=\frac{\mathrm{c}}{\mathrm{sin}\mathrm{C}}=\mathrm{k}\left(\mathrm{let}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}\mathrm{B}+\mathrm{sin}\mathrm{C}}{\mathrm{sin}\mathrm{A}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2\mathrm{sin}\left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)\mathrm{cos}\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{2\mathrm{sin}\frac{\mathrm{A}}{2}\mathrm{cos}\frac{\mathrm{A}}{2}}\left[\begin{array}{l}\mathrm{sin}\mathrm{x}+\mathrm{sin}\mathrm{y}\\ =2\mathrm{sin}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\mathrm{cos}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{A}}{2}\right)\mathrm{cos}\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{\mathrm{sin}\frac{\mathrm{A}}{2}\mathrm{cos}\frac{\mathrm{A}}{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{cos}\frac{\mathrm{A}}{2}\mathrm{cos}\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{B}+\mathrm{C}}{2}\right)\mathrm{cos}\left(\frac{\mathrm{A}}{2}\right)}\left[\because \mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)=\mathrm{cos\theta }\right]\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{cos}\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{\mathrm{cos}\left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{b}+\mathrm{c}\right)\mathrm{cos}\text{}\frac{\mathrm{B}+\mathrm{C}}{2}\text{}=\mathrm{a}\mathrm{cos}\frac{\mathrm{B}-\mathrm{C}}{2}\text{is proved.}\end{array}$

Q.52 For any triangle ABC, prove that a cos A + b cos B + c cos C = 2 a sin B sin C

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\text{a cos A}+\text{b cos B}+\text{c cos C}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{k}\mathrm{sin}\mathrm{A}\mathrm{cos}\mathrm{A}+\mathrm{k}\mathrm{sin}\mathrm{B}\mathrm{cos}\mathrm{B}+\mathrm{k}\mathrm{sin}\mathrm{C}\mathrm{cos}\mathrm{C}\\ \left[\because \frac{\mathrm{a}}{\mathrm{sinA}}=\frac{\mathrm{b}}{\mathrm{sinB}}=\frac{\mathrm{c}}{\mathrm{sinC}}=\mathrm{k}\left(\mathrm{let}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}}{2}\left(2\mathrm{sin}\mathrm{A}\mathrm{cos}\mathrm{A}+2\mathrm{sin}\mathrm{B}\mathrm{cos}\mathrm{B}+2\mathrm{sin}\mathrm{C}\mathrm{cos}\mathrm{C}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}}{2}\left(\mathrm{sin}2\mathrm{A}+\mathrm{sin}2\mathrm{B}+\mathrm{sin}2\mathrm{C}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}}{2}\left(2\mathrm{sin}\frac{2\mathrm{A}+2\mathrm{B}}{2}.\mathrm{cos}\frac{2\mathrm{A}-2\mathrm{B}}{2}+\mathrm{sin}2\mathrm{C}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}}{2}\left\{2\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right).\mathrm{cos}\left(\mathrm{A}-\mathrm{B}\right)+\mathrm{sin}2\mathrm{C}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}}{2}\left\{2\mathrm{sin}\left(\mathrm{\pi }-\mathrm{C}\right).\mathrm{cos}\left(\mathrm{A}-\mathrm{B}\right)+2\mathrm{sin}\mathrm{C}\mathrm{cos}\mathrm{C}\right\}\\ \left[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{\pi }\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}}{2}\left\{2\mathrm{sin}\mathrm{C}.\mathrm{cos}\left(\mathrm{A}-\mathrm{B}\right)+2\mathrm{sin}\mathrm{C}\mathrm{cos}\left(\mathrm{\pi }-\mathrm{A}-\mathrm{B}\right)\right\}\\ \text{\hspace{0.17em}}\left[\because \mathrm{sin}\left(\mathrm{\pi }-\mathrm{\theta }\right)=\mathrm{sin\theta }\right]\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{k}\mathrm{sin}\mathrm{C}\left\{\mathrm{cos}\left(\mathrm{A}-\mathrm{B}\right)-\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{k}\mathrm{sin}\mathrm{C}\left[2\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}+\mathrm{A}+\mathrm{B}}{2}\right)\mathrm{sin}\left\{\frac{\mathrm{A}+\mathrm{B}-\left(\mathrm{A}-\mathrm{B}\right)}{2}\right\}\right]\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2\mathrm{k}\mathrm{sin}\mathrm{C}.\mathrm{sin}\mathrm{A}.\mathrm{sin}\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2\left(\mathrm{k}\mathrm{sin}\mathrm{A}\right).\mathrm{sin}\mathrm{B}.\mathrm{sin}\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2\mathrm{a}\mathrm{sin}\mathrm{B}.\mathrm{sin}\mathrm{C}\left[\frac{\mathrm{sin}\mathrm{A}}{\mathrm{a}}=\mathrm{k}\left(\mathrm{let}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{it is proved.}\end{array}$

Q.53

$\begin{array}{l}\text{For any triangle ABC},\text{prove that}\\ \frac{\mathrm{cos}\text{\hspace{0.17em}}\mathrm{A}}{\mathrm{a}}+\frac{\mathrm{cos}\text{\hspace{0.17em}}\mathrm{B}}{\mathrm{b}}+\frac{\mathrm{cos}\text{\hspace{0.17em}}\mathrm{C}}{\mathrm{c}}=\frac{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}}{2\mathrm{abc}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{cos}\text{\hspace{0.17em}}\mathrm{A}}{\mathrm{a}}+\frac{\mathrm{cos}\text{\hspace{0.17em}}\mathrm{B}}{\mathrm{b}}+\frac{\mathrm{cos}\text{\hspace{0.17em}}\mathrm{C}}{\mathrm{c}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{a}}\left(\frac{{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-{\mathrm{a}}^{2}}{2\mathrm{bc}}\right)+\frac{1}{\mathrm{b}}\left(\frac{{\mathrm{c}}^{2}+{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}{2\mathrm{ca}}\right)+\frac{1}{\mathrm{c}}\left(\frac{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{2\mathrm{ab}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-{\mathrm{a}}^{2}+{\mathrm{c}}^{2}+{\mathrm{a}}^{2}-{\mathrm{b}}^{2}+{\mathrm{a}}^{2}+{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{2\mathrm{abc}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{b}}^{2}+{\mathrm{c}}^{2}+{\mathrm{a}}^{2}}{2\mathrm{abc}}\\ \mathrm{Hence},\text{it is proved.}\end{array}$

Q.54 For any triangle ABC, prove that
(b2 – c2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0

Ans.

$\left({\text{b}}^{\text{2}}-{\text{c}}^{\text{2}}\right)\text{cot A}=\left({\text{b}}^{\text{2}}-{\text{c}}^{\text{2}}\right)\frac{\mathrm{cos}\mathrm{A}}{\mathrm{sin}\mathrm{A}}$ $\begin{array}{l}=\left({\text{b}}^{\text{2}}-{\text{c}}^{\text{2}}\right)\frac{\mathrm{cos}\mathrm{A}}{\mathrm{k}\text{\hspace{0.17em}}\mathrm{a}}\left[\because \frac{\mathrm{sin}\mathrm{A}}{\mathrm{a}}=\frac{\mathrm{sin}\mathrm{B}}{\mathrm{b}}=\frac{\mathrm{sin}\mathrm{C}}{\mathrm{c}}=\mathrm{k}\right]\\ =\frac{\left({\text{b}}^{\text{2}}\text{}-{\text{c}}^{\text{2}}\right)}{\mathrm{ka}}×\frac{{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-{\mathrm{a}}^{2}}{2\mathrm{bc}}\\ =\frac{{\mathrm{b}}^{4}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}-{\mathrm{b}}^{2}{\mathrm{a}}^{2}-{\mathrm{b}}^{2}{\mathrm{c}}^{2}-{\mathrm{c}}^{4}+{\mathrm{c}}^{2}{\mathrm{a}}^{2}}{2\mathrm{kabc}}\\ \left({\text{c}}^{\text{2}}\text{}-{\text{a}}^{\text{2}}\right)\text{cot B}=\left({\text{c}}^{\text{2}}\text{}-{\text{a}}^{\text{2}}\right)\frac{\mathrm{cos}\mathrm{B}}{\mathrm{sin}\mathrm{B}}\\ =\left({\text{c}}^{\text{2}}\text{}-{\text{a}}^{\text{2}}\right)\frac{\mathrm{cos}\mathrm{B}}{\mathrm{k}\text{\hspace{0.17em}}\mathrm{b}}\\ =\frac{\left({\text{c}}^{\text{2}}\text{}-{\text{a}}^{\text{2}}\right)}{\mathrm{kb}}×\frac{{\mathrm{c}}^{2}+{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}{2\mathrm{bc}}\\ =\frac{{\mathrm{c}}^{4}+{\mathrm{c}}^{2}{\mathrm{a}}^{2}-{\mathrm{c}}^{2}{\mathrm{b}}^{2}-{\mathrm{a}}^{2}{\mathrm{c}}^{2}-{\mathrm{a}}^{4}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}{2\mathrm{kabc}}\\ \left({\text{a}}^{\text{2}}\text{}-{\text{b}}^{\text{2}}\right)\text{cot C}=\left({\text{a}}^{\text{2}}\text{}-{\text{b}}^{\text{2}}\right)\frac{\mathrm{cos}\mathrm{C}}{\mathrm{sin}\mathrm{C}}\\ =\left({\text{a}}^{\text{2}}\text{}-{\text{b}}^{\text{2}}\right)\frac{\mathrm{cos}\mathrm{C}}{\mathrm{k}\text{\hspace{0.17em}}\mathrm{c}}\\ =\frac{\left({\text{a}}^{\text{2}}\text{}-{\text{b}}^{\text{2}}\right)}{\mathrm{kc}}×\frac{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{2\mathrm{ab}}\\ =\frac{{\mathrm{a}}^{4}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}-{\mathrm{a}}^{2}{\mathrm{c}}^{2}-{\mathrm{a}}^{2}{\mathrm{b}}^{2}-{\mathrm{b}}^{4}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}}{2\mathrm{kabc}}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\left({\mathrm{b}}^{2}-{\mathrm{c}}^{2}\right)\mathrm{cot}\mathrm{A}+\left({\mathrm{c}}^{2}-{\mathrm{a}}^{2}\right)\mathrm{cot}\mathrm{B}+\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)\mathrm{cot}\mathrm{C}\\ \text{\hspace{0.17em}}=\frac{{\mathrm{b}}^{4}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}-{\mathrm{b}}^{2}{\mathrm{a}}^{2}-{\mathrm{b}}^{2}{\mathrm{c}}^{2}-{\mathrm{c}}^{4}+{\mathrm{c}}^{2}{\mathrm{a}}^{2}}{2\mathrm{kabc}}\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}+\frac{{\mathrm{c}}^{4}+{\mathrm{c}}^{2}{\mathrm{a}}^{2}-{\mathrm{c}}^{2}{\mathrm{b}}^{2}-{\mathrm{a}}^{2}{\mathrm{c}}^{2}-{\mathrm{a}}^{4}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}{2\mathrm{kabc}}+\frac{{\mathrm{a}}^{4}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}-{\mathrm{a}}^{2}{\mathrm{c}}^{2}-{\mathrm{a}}^{2}{\mathrm{b}}^{2}-{\mathrm{b}}^{4}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}}{2\mathrm{kabc}}\\ \text{\hspace{0.17em}}=\frac{\left(\begin{array}{l}{\mathrm{b}}^{4}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}-{\mathrm{b}}^{2}{\mathrm{a}}^{2}-{\mathrm{b}}^{2}{\mathrm{c}}^{2}-{\mathrm{c}}^{4}+{\mathrm{c}}^{2}{\mathrm{a}}^{2}+{\mathrm{c}}^{4}+{\mathrm{c}}^{2}{\mathrm{a}}^{2}-{\mathrm{c}}^{2}{\mathrm{b}}^{2}\\ -{\mathrm{a}}^{2}{\mathrm{c}}^{2}-{\mathrm{a}}^{4}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\mathrm{a}}^{4}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}-{\mathrm{a}}^{2}{\mathrm{c}}^{2}-{\mathrm{a}}^{2}{\mathrm{b}}^{2}-{\mathrm{b}}^{4}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}\end{array}\right)}{2\mathrm{kabc}}\\ \text{\hspace{0.17em}}=\frac{0}{2\mathrm{kabc}}=0\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\mathrm{Thus},\text{it is proved.}\end{array}$

Q.55

$\begin{array}{l}\text{For any triangle ABC},\text{prove that}\\ \frac{{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}}\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{A}+\frac{{\mathrm{c}}^{2}-{\mathrm{a}}^{2}}{{\mathrm{b}}^{2}}\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{B}+\frac{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}{{\mathrm{c}}^{2}}\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{C}=0\end{array}$

Ans.

$\begin{array}{l}\frac{{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}}\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{A}\text{\hspace{0.17em}}=\left(\frac{{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}}\right)2\mathrm{sin}\text{\hspace{0.17em}}\mathrm{A}\mathrm{cos}\mathrm{A}\\ =\left(\frac{{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}}\right)2\left(\mathrm{k}\text{\hspace{0.17em}}\mathrm{a}\right)\mathrm{cosA}\left[\because \frac{\mathrm{sin}\mathrm{A}}{\mathrm{a}}=\frac{\mathrm{sin}\mathrm{B}}{\mathrm{b}}=\frac{\mathrm{sin}\mathrm{C}}{\mathrm{c}}=\mathrm{k}\right]\\ =\left(\frac{{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}}\right)2\left(\mathrm{k}\text{\hspace{0.17em}}\mathrm{a}\right)\frac{{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-{\mathrm{a}}^{2}}{2\mathrm{bc}}\\ =\mathrm{k}\left(\frac{{\mathrm{b}}^{4}-{\mathrm{c}}^{4}-{\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}}{2\mathrm{abc}}\right)\\ \mathrm{Similarly},\\ \frac{{\mathrm{c}}^{2}-{\mathrm{a}}^{2}}{{\mathrm{b}}^{2}}\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{B}=\mathrm{k}\left(\frac{{\mathrm{c}}^{4}-{\mathrm{a}}^{4}-{\mathrm{c}}^{2}{\mathrm{b}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}{2\mathrm{abc}}\right)\text{and}\end{array}$ $\begin{array}{l}\frac{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}{{\mathrm{c}}^{2}}\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{C}=\mathrm{k}\left(\frac{{\mathrm{a}}^{4}-{\mathrm{b}}^{4}-{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}}{2\mathrm{abc}}\right)\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}}\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{A}+\frac{{\mathrm{c}}^{2}-{\mathrm{a}}^{2}}{{\mathrm{b}}^{2}}\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{B}+\frac{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}{{\mathrm{c}}^{2}}\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{C}\\ =\mathrm{k}\left(\frac{{\mathrm{b}}^{4}-{\mathrm{c}}^{4}-{\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}}{2\mathrm{abc}}\right)+\mathrm{k}\left(\frac{{\mathrm{c}}^{4}-{\mathrm{a}}^{4}-{\mathrm{c}}^{2}{\mathrm{b}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}{2\mathrm{abc}}\right)\\ +\mathrm{k}\left(\frac{{\mathrm{a}}^{4}-{\mathrm{b}}^{4}-{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}}{2\mathrm{abc}}\right)\\ =\mathrm{k}\left(\frac{\begin{array}{l}{\mathrm{b}}^{4}-{\mathrm{c}}^{4}-{\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{c}}^{4}-{\mathrm{a}}^{4}-{\mathrm{c}}^{2}{\mathrm{b}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}\\ +{\mathrm{a}}^{4}-{\mathrm{b}}^{4}-{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}\end{array}}{2\mathrm{abc}}\right)\\ =\mathrm{k}\left(\frac{0}{2\mathrm{abc}}\right)\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Hence proved.}\end{array}$

Q.56 A tree stands vertically on a hill side which makes an angle of 15° with the horizontal. From a point on the ground 35 m down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°. Find the height of the tree.

Ans.

Let PQ be a tree of height h m and A be a point on ground as AQ = 35 m.

$\begin{array}{l}\angle \mathrm{APR}=90\mathrm{°}-60\mathrm{°}\\ =30\mathrm{°}\end{array}$ $\begin{array}{l}⇒\angle \mathrm{APQ}=30\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PAQ}=60\mathrm{°}-15\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=45\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PQA}=180\mathrm{°}-\angle \mathrm{PAQ}-\angle \mathrm{APQ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=180\mathrm{°}-45\mathrm{°}-30\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=105\mathrm{°}\\ \mathrm{Applying}\text{sin rule in}\mathrm{\Delta PQA},\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{sin}45\mathrm{°}}{\mathrm{PQ}}=\frac{\mathrm{sin}30\mathrm{°}}{\mathrm{AQ}}=\frac{\mathrm{sin}105\mathrm{°}}{\mathrm{AP}}\\ ⇒\frac{\mathrm{sin}45\mathrm{°}}{\mathrm{PQ}}=\frac{\mathrm{sin}30\mathrm{°}}{35}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\left(\frac{1}{\sqrt{2}}\right)}{\mathrm{PQ}}=\frac{\left(\frac{1}{2}\right)}{35}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\frac{35}{\sqrt{2}}=\frac{\mathrm{PQ}}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}=35\sqrt{2}\\ \mathrm{Thus},\text{the height of the tree is 35}\sqrt{2}\text{\hspace{0.17em}}\mathrm{m}.\end{array}$

Q.57 Two ships leave a port at the same time. One goes 24 km per hour in the direction N45°E and other travels 32 km per hour in the direction S75°E. Find the distance between the ships at the end of 3 hours.

Ans.

Speed of first ship = 24 km/hr
Speed of second ship = 32 km/hr
Distance (OP) covered in 3 hours
= 24 × 3
= 72 km

Distance (OQ) covered in 3 hours
= 32 × 3
= 96 km
Angle POQ = (90° – 45°) + (90° – 75°)
= 45° + 15°
= 60°

$\begin{array}{l}\mathrm{By}\text{using cosine formula in}\mathrm{\Delta OPQ},\text{we get}\\ {\mathrm{PQ}}^{2}={\mathrm{OP}}^{2}+{\mathrm{OQ}}^{2}-2×\text{\hspace{0.17em}}\mathrm{OP}×\mathrm{OQ}\mathrm{cos}60\mathrm{°}\\ ={\left(72\right)}^{2}+{\left(96\right)}^{2}-2×72×96×\frac{1}{2}\\ =5184+9216-6912\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}=\sqrt{7488}\\ =86.53\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{km}\end{array}$

Q.58

$\begin{array}{l}\mathrm{Two}\mathrm{trees},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{a}\mathrm{river}.\mathrm{From}\mathrm{a}\\ \mathrm{point}\mathrm{C}\mathrm{in}\mathrm{the}\mathrm{river}\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{trees}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{is}250\mathrm{m}\\ \mathrm{and}300\mathrm{m},\mathrm{respectively}.\mathrm{If}\mathrm{the}\mathrm{angle}\mathrm{C}\mathrm{is}45\mathrm{°},\mathrm{find}\mathrm{the}\\ \mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{trees}\left(\mathrm{use}\sqrt{2}=1.44\right).\end{array}$

Ans.

Let A and B be the positions of two trees respectively. The distance of first tree from C is 250m and that of second tree is 300 m.
Angle C is 45°.

$\begin{array}{l}\mathrm{Applying}\text{cosine formula in}\mathrm{\Delta ABC},\text{}\mathrm{we}\text{}\mathrm{get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos}\mathrm{C}=\frac{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}-{\mathrm{c}}^{2}}{2\mathrm{ab}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos}\text{}45\mathrm{°}=\frac{{\left(300\right)}^{2}+{\left(250\right)}^{2}-{\mathrm{c}}^{2}}{2\left(300\right)\left(250\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{\sqrt{2}}=\frac{90000+62500-{\mathrm{c}}^{2}}{150000}\end{array}$ $\begin{array}{l}\frac{150000}{\sqrt{2}}×\frac{\sqrt{2}}{\sqrt{2}}=152500-{\mathrm{AB}}^{2}\\ \text{ }\frac{150000\sqrt{2}}{2}=152500-{\mathrm{AB}}^{2}\\ \frac{150000×1.414}{2}=152500-{\mathrm{AB}}^{2}\\ \text{ }106050=152500-{\mathrm{AB}}^{2}\\ \text{ }\mathrm{AB}=\sqrt{152500-106050}\\ \text{ }=\sqrt{46450}\\ \text{ }\mathrm{AB}=215.5\text{ m}\\ \mathrm{Thus},\text{distance between two trees is 215.5m.}\end{array}$

Q.59 Prove that:

$2\mathrm{cos}\frac{\mathrm{\pi }}{13}\mathrm{cos}\frac{9\mathrm{\pi }}{13}+\mathrm{cos}\frac{3\mathrm{\pi }}{13}+\mathrm{cos}\frac{5\mathrm{\pi }}{13}=0$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=2\mathrm{cos}\frac{\mathrm{\pi }}{13}\mathrm{cos}\frac{9\mathrm{\pi }}{13}+\mathrm{cos}\frac{3\mathrm{\pi }}{13}+\mathrm{cos}\frac{5\mathrm{\pi }}{13}\\ =\mathrm{cos}\left(\frac{\mathrm{\pi }}{13}+\frac{9\mathrm{\pi }}{13}\right)+\mathrm{cos}\left(\frac{\mathrm{\pi }}{13}-\frac{9\mathrm{\pi }}{13}\right)+\mathrm{cos}\left(\mathrm{\pi }-\frac{10\mathrm{\pi }}{13}\right)+\mathrm{cos}\left(\mathrm{\pi }-\frac{8\mathrm{\pi }}{13}\right)\\ =\mathrm{cos}\left(\frac{10\mathrm{\pi }}{13}\right)+\mathrm{cos}\left(-\frac{8\mathrm{\pi }}{13}\right)-\mathrm{cos}\frac{10\mathrm{\pi }}{13}-\mathrm{cos}\frac{8\mathrm{\pi }}{13}\end{array}$ $\begin{array}{l}=\mathrm{cos}\left(\frac{8\mathrm{\pi }}{13}\right)-\mathrm{cos}\frac{8\mathrm{\pi }}{13}\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.60 Prove that:
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\left(\mathrm{sin}3\mathrm{x}+\mathrm{sin}\mathrm{x}\right)\mathrm{sinx}+\left(\mathrm{cos}3\mathrm{x}-\mathrm{cos}\mathrm{x}\right)\mathrm{cos}\mathrm{x}\\ \text{\hspace{0.17em}}=\mathrm{sin}3\mathrm{x}\mathrm{sin}\mathrm{x}+{\mathrm{sin}}^{2}\mathrm{x}+\mathrm{cos}3\mathrm{x}\mathrm{cos}\mathrm{x}-{\mathrm{cos}}^{2}\mathrm{x}\\ \text{\hspace{0.17em}}=\mathrm{cos}3\mathrm{x}\mathrm{cos}\mathrm{x}+\mathrm{sin}3\mathrm{x}\mathrm{sin}\mathrm{x}-{\mathrm{cos}}^{2}\mathrm{x}+{\mathrm{sin}}^{2}\mathrm{x}\\ \text{\hspace{0.17em}}=\mathrm{cos}\left(3\mathrm{x}-\mathrm{x}\right)-\left({\mathrm{cos}}^{2}\mathrm{x}-{\mathrm{sin}}^{2}\mathrm{x}\right)\\ \left[\because \mathrm{cos}\left(\mathrm{A}-\mathrm{B}\right)=\mathrm{cos}\mathrm{A}\mathrm{cos}\mathrm{B}+\mathrm{sin}\mathrm{A}\mathrm{cos}\mathrm{B}\right]\\ \text{\hspace{0.17em}}=\mathrm{cos}2\mathrm{x}-\mathrm{cos}2\mathrm{x}\left[\because {\mathrm{cos}}^{2}\mathrm{x}-{\mathrm{sin}}^{2}\mathrm{x}=\mathrm{cos}2\mathrm{x}\right]\\ \text{\hspace{0.17em}}=0\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.61 Prove that:

${\left(\mathrm{cos}\mathrm{x}+\mathrm{cos}\mathrm{y}\right)}^{2}+{\left(\mathrm{sin}\mathrm{x}-\mathrm{sin}\mathrm{y}\right)}^{2}=4{\mathrm{cos}}^{2}\frac{\mathrm{x}+\mathrm{y}}{2}$

Ans.

$\begin{array}{l}{\text{L.H.S. : (cos x + cos y)}}^{\text{2}}{\text{+ (sin x-sin y)}}^{\text{2}}\text{}\\ \text{=}{\left(\text{2cos}\frac{\text{x + y}}{\text{2}}\text{cos}\frac{\text{x – y}}{\text{2}}\right)}^{\text{2}}\text{+}{\left(\text{2cos}\frac{\text{x + y}}{\text{2}}\text{sin}\frac{\text{x}–\text{y}}{\text{2}}\right)}^{\text{2}}\\ {\text{= 4cos}}^{\text{2}}\text{}\left(\frac{\text{x + y}}{\text{2}}\right)\left({\text{cos}}^{\text{2}}\text{}\frac{\text{x}–\text{y}}{\text{2}}{\text{+sin}}^{\text{2}}\text{}\frac{\text{x}–\text{y}}{\text{2}}\right)\\ {\text{= 4 cos}}^{\text{2}}\left(\frac{\text{x + y}}{\text{2}}\right)\text{× 1}\left[{\text{Q sin}}^{\text{2}}{\text{x + cos}}^{\text{2}}\text{x = 1}\right]\end{array}$ $\begin{array}{l}=4{\mathrm{cos}}^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.62 Prove that:

${\left(\mathrm{cos}\mathrm{x}-\mathrm{cos}\mathrm{y}\right)}^{2}+{\left(\mathrm{sin}\mathrm{x}-\mathrm{sin}\mathrm{y}\right)}^{2}=4{\mathrm{sin}}^{2}\frac{\mathrm{x}-\mathrm{y}}{2}$

Ans.

$\begin{array}{l}{\text{L.H.S. : (cos x – cos y)}}^{\text{2}}\text{+ (sin x}–{\text{sin y)}}^{\text{2}}\\ \text{=}{\left\{–\text{2sin}\left(\frac{\text{x + y}}{\text{2}}\right)\text{sin}\left(\frac{\text{x}–\text{y}}{\text{2}}\right)\right\}}^{\text{2}}\text{+}{\left\{\text{2cos}\left(\frac{\text{x + y}}{\text{2}}\right)\text{sin}\left(\text{}\frac{\text{x}–\text{y}}{\text{2}}\right)\right\}}^{\text{2}}\\ {\text{=4sin}}^{\text{2}}\text{}\left(\frac{\text{x}–\text{y}}{\text{2}}\right)\left\{{\text{sin}}^{\text{2}}\left(\frac{\text{x + y}}{\text{2}}\right){\text{+cos}}^{\text{2}}\text{}\left(\frac{\text{x + y}}{\text{2}}\right)\right\}\\ {\text{= 4 sin}}^{\text{2}}\text{}\left(\frac{\text{x}–\text{y}}{\text{2}}\right)\text{× 1}\left[{\text{Q sin}}^{\text{2}}{\text{x + cos}}^{\text{2}}\text{x = 1}\right]\end{array}$ $\begin{array}{l}=4{\mathrm{sin}}^{2}\left(\frac{\mathrm{x}–\mathrm{y}}{2}\right)\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.63 Prove that:
sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Ans.

$\begin{array}{l}\text{L.H.S.}=\text{sin x}+\text{sin 3x}+\text{sin 5x}+\text{sin 7x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{sin 7x}+\text{sin x}+\text{sin 5x}+\text{sin 3x}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{sin}\left(\frac{7\mathrm{x}+\mathrm{x}}{2}\right)\mathrm{cos}\left(\frac{7\mathrm{x}-\mathrm{x}}{2}\right)+2\mathrm{sin}\left(\frac{5\mathrm{x}+3\mathrm{x}}{2}\right)\mathrm{cos}\left(\frac{5\mathrm{x}-3\mathrm{x}}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{sin}4\mathrm{x}\text{\hspace{0.17em}}\mathrm{cos}\text{}3\mathrm{x}+2\mathrm{sin}4\mathrm{x}\mathrm{cos}\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{sin}4\mathrm{x}\left(\mathrm{cos}3\mathrm{x}+\mathrm{cos}\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{sin}4\mathrm{x}\left\{2\mathrm{cos}\left(\frac{3\mathrm{x}+\mathrm{x}}{2}\right)\mathrm{cos}\left(\frac{3\mathrm{x}-\mathrm{x}}{2}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\mathrm{sin}4\mathrm{x}\mathrm{cos}2\mathrm{x}\mathrm{cos}\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\mathrm{cos}\mathrm{x}\mathrm{cos}2\mathrm{x}\mathrm{sin}4\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.64

Prove that:

$\frac{\left(\mathrm{sin}7\mathrm{x}+\mathrm{sin}5\mathrm{x}\right)+\left(\mathrm{sin}9\mathrm{x}+\mathrm{sin}3\mathrm{x}\right)}{\left(\mathrm{cos}7\mathrm{x}+\mathrm{cos}5\mathrm{x}\right)+\left(\mathrm{cos}9\mathrm{x}+\mathrm{cos}3\mathrm{x}\right)}=\mathrm{tan}6\mathrm{x}$

Ans.

$\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\left(\mathrm{sin}7\mathrm{x}+\mathrm{sin}5\mathrm{x}\right)+\left(\mathrm{sin}9\mathrm{x}+\mathrm{sin}3\mathrm{x}\right)}{\left(\mathrm{cos}7\mathrm{x}+\mathrm{cos}5\mathrm{x}\right)+\left(\mathrm{cos}9\mathrm{x}+\mathrm{cos}3\mathrm{x}\right)}$ $\begin{array}{l}\text{\hspace{0.17em}}=\frac{\left(2\mathrm{sin}\frac{7\mathrm{x}+5\mathrm{x}}{2}\mathrm{cos}\frac{7\mathrm{x}-5\mathrm{x}}{2}\right)+\left(2\mathrm{sin}\frac{9\mathrm{x}+3\mathrm{x}}{2}\mathrm{cos}\frac{9\mathrm{x}-3\mathrm{x}}{2}\right)}{\left(2\mathrm{cos}\frac{7\mathrm{x}+5\mathrm{x}}{2}\mathrm{cos}\frac{7\mathrm{x}-5\mathrm{x}}{2}\right)+\left(2\mathrm{cos}\frac{9\mathrm{x}+3\mathrm{x}}{2}\mathrm{cos}\frac{9\mathrm{x}-3\mathrm{x}}{2}\right)}\\ \text{\hspace{0.17em}}=\frac{\left(2\mathrm{sin}6\mathrm{x}\mathrm{cos}\mathrm{x}\right)+\left(2\mathrm{sin}6\mathrm{x}\mathrm{cos}3\mathrm{x}\right)}{\left(2\mathrm{cos}6\mathrm{x}\mathrm{cos}\mathrm{x}\right)+\left(2\mathrm{cos}6\mathrm{x}\mathrm{cos}3\mathrm{x}\right)}\\ \text{\hspace{0.17em}}=\frac{2\mathrm{sin}6\mathrm{x}\left(\mathrm{cos}\mathrm{x}+\mathrm{cos}3\mathrm{x}\right)}{2\mathrm{cos}6\mathrm{x}\left(\mathrm{cos}\mathrm{x}+\mathrm{cos}3\mathrm{x}\right)}\\ \text{\hspace{0.17em}}=\mathrm{tan}6\mathrm{x}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\mathrm{proved}.\end{array}$

Q.65

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \mathrm{sin}3\mathrm{x}+\mathrm{sin}2\mathrm{x}-\mathrm{sinx}=4\mathrm{sinxcos}\frac{\mathrm{x}}{2}\text{\hspace{0.17em}}\mathrm{cos}\frac{3\mathrm{x}}{2}\end{array}$

Ans.

$\begin{array}{l}\text{L.H.S. = sin 3x+sin 2x}–\text{sin x}\\ \text{ = sin 3x}–\text{sin x + sin 2x}\\ \text{ = 2cos}\left(\frac{\text{3x + x}}{\text{2}}\right)\text{sin}\left(\frac{\text{3x}–\text{x}}{\text{2}}\right)\text{+2sin x cos x}\\ \text{ = 2cos 2x sin x + 2sin x cos x}\\ \text{ = 2sin x}\left(\text{cos 2x + cos x}\right)\\ \text{ = 2sin x}\left\{\text{2cos}\left(\frac{\text{2x + x}}{\text{2}}\right)\text{cos}\left(\frac{\text{2x}–\text{x}}{\text{2}}\right)\right\}\\ \text{ = 4sin x cos}\frac{\text{3x}}{\text{2}}\text{ cos}\frac{\text{x}}{\text{2}}\\ \text{ = 4sin x cos}\frac{\text{x}}{\text{2}}\text{ cos}\frac{\text{3x}}{\text{2}}\\ \text{ = R.H.S. Hence proved.}\end{array}$

Q.66

$\mathrm{Find} \mathrm{sin}\frac{\mathrm{x}}{2},\mathrm{cos}\frac{\mathrm{x}}{2}\mathrm{and}\mathrm{tan}\frac{\mathrm{x}}{2}\mathrm{in}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}:$ $8.\mathrm{t}\mathrm{an}\mathrm{x}=-\frac{4}{3},\mathrm{x}\mathrm{in}\mathrm{quadrant}\mathrm{II}$ $9.\mathrm{c}\mathrm{os}\mathrm{x}=-\frac{1}{3},\mathrm{x}\mathrm{in}\mathrm{quadrant}\mathrm{III}$ $10.\text{ }\mathrm{sinx}=\frac{1}{4},\mathrm{x}\text{ in quadrant II}$

Ans.

$\begin{array}{l}8.\\ \mathrm{Since},\text{ }\mathrm{tan}\text{ }\mathrm{x}=-\frac{4}{3},\mathrm{xinquadrantII}\\ \because \text{ }{\mathrm{sec}}^{2}\text{x}=1+{\mathrm{tan}}^{2}\text{x}\\ \text{ }=1+{\left(-\frac{4}{3}\right)}^{2}\\ \text{ }=1+\frac{16}{9}\\ \text{ }=\frac{25}{9}\\ \text{ }\mathrm{sec}\text{ }\mathrm{x}=±\sqrt{\frac{25}{9}}\\ \text{ }=-\frac{5}{3}\text{ }\left[\text{x is in II quadrant.}\right]\\ ⇒\text{ }\mathrm{cos}\text{ }\mathrm{x}=-\frac{3}{5}\\ \text{ }2{\mathrm{cos}}^{2}\frac{\text{x}}{2}-1=-\frac{3}{5}\\ \text{ }2{\mathrm{cos}}^{2}\frac{\text{x}}{2}=1-\frac{3}{5}\\ \text{ }=\frac{2}{5}\\ \text{ }{\mathrm{cos}}^{2}\frac{\text{x}}{2}=\frac{1}{5}\\ \text{ }\mathrm{cos}\frac{\text{x}}{2}=±\frac{1}{\sqrt{5}}\\ \mathrm{cos}\frac{\text{x}}{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{l}\because \text{x }\mathrm{lies}\text{ in II quadrant,}\\ \text{So,}\frac{\text{x}}{2}\text{ lies in I quadrant.}\end{array}\right]\\ \text{ }=\frac{1}{\sqrt{5}}×\frac{\sqrt{5}}{\sqrt{5}}\\ \text{ }=\frac{\sqrt{5}}{5}\\ \because \text{ }\mathrm{sin}\frac{\text{x}}{2}=\sqrt{1-{\mathrm{cos}}^{2}\frac{\text{x}}{2}}\\ \text{ }=±\sqrt{1-{\left(\frac{1}{\sqrt{5}}\right)}^{2}}\\ \text{ }=±\sqrt{1-\frac{1}{5}}\\ \text{ }=\frac{2}{\sqrt{5}}\left[\because \frac{\text{x}}{2}\text{ lies in I quadrant.}\right]\\ \text{ }=\frac{2}{\sqrt{5}}×\frac{\sqrt{5}}{\sqrt{5}}\\ \text{ }=\frac{2\sqrt{5}}{5}\\ \text{ }\mathrm{tan}\frac{\text{x}}{2}=\frac{\mathrm{sin}\left(\frac{\text{x}}{2}\right)}{\mathrm{cos}\left(\frac{\text{x}}{2}\right)}\\ \text{ }=\frac{\left(\frac{2}{\sqrt{5}}\right)}{\left(\frac{1}{\sqrt{5}}\right)}=2\\ \mathrm{So},\text{the values of sin}\frac{\text{x}}{2},\text{cos}\frac{\text{x}}{2}\text{and tan}\frac{\text{x}}{2}\text{are}\frac{\sqrt{5}}{5},\text{ }\frac{2\sqrt{5}}{5}\text{and}\\ \text{2 respectively.}\end{array}$ $\begin{array}{l}9.\\ \mathrm{We}\text{have,}\\ \text{ }\mathrm{cosx}=-\frac{1}{3}\\ 2{\mathrm{cos}}^{2}\frac{\text{x}}{2}-1=-\frac{1}{3}\\ \text{ }2{\mathrm{cos}}^{2}\frac{\text{x}}{2}=1-\frac{1}{3}\\ \text{ }=\frac{2}{3}\\ \text{ }{\mathrm{cos}}^{2}\frac{\text{x}}{2}=\frac{1}{3}\\ \text{ }\mathrm{cos}\frac{\text{x}}{2}=±\frac{1}{\sqrt{3}}\\ \mathrm{cos}\frac{\text{x}}{2}=-\frac{1}{\sqrt{3}}\left[\begin{array}{l}\because \text{x }\mathrm{lies}\text{ in III quadrant,}\\ \text{So,}\frac{\text{x}}{2}\text{ lies in II quadrant.}\end{array}\right]\\ \text{ }=-\frac{1}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\\ \text{ }=-\frac{\sqrt{3}}{3}\\ \because \text{ }\mathrm{sin}\frac{\text{x}}{2}=\sqrt{1-{\mathrm{cos}}^{2}\frac{\text{x}}{2}}\\ \text{ }=±\sqrt{1-{\left(-\frac{1}{\sqrt{3}}\right)}^{2}}\\ \text{ }=±\sqrt{1-\frac{1}{3}}\\ \text{ }=\frac{\sqrt{2}}{\sqrt{3}}\left[\because \text{ }\frac{\text{x}}{2}\text{ lies in II quadrant.}\right]\\ \text{ }=\frac{\sqrt{2}}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\\ \text{ }=\frac{\sqrt{6}}{3}\\ \text{ }\mathrm{tan}\frac{\text{x}}{2}=\frac{\mathrm{sin}\left(\frac{\text{x}}{2}\right)}{\mathrm{cos}\left(\frac{\text{x}}{2}\right)}\\ \text{ }=\frac{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}{\left(-\frac{1}{\sqrt{3}}\right)}=-\sqrt{2}\\ \mathrm{So},\text{the values of sin}\frac{\text{x}}{2},\text{cos}\frac{\text{x}}{2}\text{and tan}\frac{\text{x}}{2}\text{are}\frac{\sqrt{6}}{3},-\frac{\sqrt{6}}{3}\text{and}\\ -\sqrt{\text{2}}\text{respectively.}\end{array}$

$\begin{array}{l}10.\\ \mathrm{We}\text{}\mathrm{have},\text{ }\mathrm{sin}\text{ }\mathrm{x}=\frac{1}{4},\text{ }\mathrm{x}\text{ }\mathrm{in}\text{ }\mathrm{quadrant}\text{ }\mathrm{II}\\ \because \text{ }\mathrm{cos}\text{ }\mathrm{x}=\sqrt{1-{\mathrm{sin}}^{2}\text{x}}\\ \text{ }\mathrm{cos}\text{ }\mathrm{x}=±\sqrt{1-{\left(\frac{1}{4}\right)}^{2}}\\ \text{ }=±\sqrt{1-\frac{1}{16}}\\ \text{ }=-\frac{\sqrt{15}}{4}\left[\because \text{ }\mathrm{x}\text{ }\mathrm{in}\text{ }\mathrm{quadrant}\text{ }\mathrm{II}\right]\\ \mathrm{Since},\text{ }\mathrm{cos}\text{ }\mathrm{x}=1-2{\mathrm{sin}}^{2}\frac{\text{x}}{2}\\ \text{ }-\frac{\sqrt{15}}{4}=1-2{\mathrm{sin}}^{2}\frac{\text{x}}{2}\\ \text{ }2{\mathrm{sin}}^{2}\frac{\text{x}}{2}=1+\frac{\sqrt{15}}{4}\\ \text{ }{\mathrm{sin}}^{2}\frac{\text{x}}{2}=\frac{1}{2}+\frac{\sqrt{15}}{8}\\ \text{ }\mathrm{sin}\frac{\text{x}}{2}=±\sqrt{\frac{1}{2}+\frac{\sqrt{15}}{8}}\\ \text{ }=\frac{\sqrt{8+2\sqrt{15}}}{4}\text{ }\left[\because \text{ }\frac{\text{x}}{2}\text{lies in I quadrant.}\right]\\ \text{ }\mathrm{and}\text{ }\mathrm{cos}\text{ }\mathrm{x}=2{\mathrm{cos}}^{2}\frac{\text{x}}{2}-1\\ -\frac{\sqrt{15}}{4}=2{\mathrm{cos}}^{2}\frac{\text{x}}{2}-1⇒{\mathrm{cos}}^{2}\frac{\text{x}}{2}=\frac{1}{2}-\frac{\sqrt{15}}{8}\\ \mathrm{cos}\frac{\text{x}}{2}=±\sqrt{\frac{1}{2}-\frac{\sqrt{15}}{8}}\\ \text{ }=\frac{\sqrt{8-2\sqrt{15}}}{4}\text{ }\left[\because \text{ }\frac{\text{x}}{2}\text{lies in I quadrant.}\right]\\ \mathrm{tan}\frac{\text{x}}{2}=\frac{\mathrm{sin}\left(\frac{\text{x}}{2}\right)}{\mathrm{cos}\left(\frac{\text{x}}{2}\right)}\\ \text{ }=\frac{\frac{\sqrt{8+2\sqrt{15}}}{4}}{\frac{\sqrt{8-2\sqrt{15}}}{4}}\\ \text{ }=\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8-2\sqrt{15}}}×\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8+2\sqrt{15}}}\\ \text{ }=\frac{8+2\sqrt{15}}{\sqrt{64-60}}\\ \text{ }=\frac{8+2\sqrt{15}}{2}\\ \mathrm{tan}\frac{\text{x}}{2}=4+\sqrt{15}\\ \mathrm{So},\text{the values of sin}\frac{\text{x}}{2},\text{cos}\frac{\text{x}}{2}\text{and tan}\frac{\text{x}}{2}\text{are}\frac{\sqrt{8+2\sqrt{15}}}{4},\\ \frac{\sqrt{8-2\sqrt{15}}}{4}\text{and}4+\sqrt{15}\text{respectively.}\end{array}$

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### 1. What trigonometric functions are provided in NCERT Solutions Class 11 Mathematics Chapter 3 ?

Trigonometric functions are real functions that are related to the angle of a right-angled triangle to the ratio of the length of its sides. Further, it discusses trigonometric identities and sum and difference identities.

### 2. What are the important formulas mentioned in NCERT Solutions Class 11 Mathematics Chapter 3?

The following are the important formulas in NCERT Solutions Class 11 Mathematics Chapter 3:

• sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
• cos(x+y) = cos(x)cos(y) – sin(x)sin(y)
• tan(x+y) = (tan x + tan y)/ (1−tan x • tan y)
• sin(x–y) = sin(x)cos(y) – cos(x)sin(y)
• cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
• tan(x−y) = (tan x–tan y)/ (1+tan x • tan y)

### 3. Do I need to attempt all questions provided in NCERT Solutions Class 11 Mathematics Chapter 3?

The questions and answers in NCERT Solutions Class 11 Mathematics Chapter 3 provide an effective way to facilitate in-depth learning of concepts. In addition, every question aims to improve the conceptual clarity of the students. Thus they can attempt long and short-form questions along with multiple-choice questions with ease.