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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.

$\begin{array}{l} Since, 1° = \frac{π}{180 °} \\ (i) 25 ° = 25 ° \times \frac{π}{180 °} \\ = \frac{5 π}{36 °} radians \\ (ii) - 47 ° 30 ‘ = - 47 .5 ° [∵ 30 ‘ = 0 .5 °] \\ = - 47 .5 ° \times \frac{π}{180 °} radians \\ = - \frac{19 π}{72} radians \\ (iii) 240 ° = 240 ° \times \frac{π}{180 °} radians \\ = \frac{4 π}{3} radians \\ (iv) 520 ° = 520 ° \times \frac{π}{180 °} radians \\ = \frac{26 π}{9} radians \end{array}$

Q.2

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

Ans.

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

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

Ans.

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

Q.4

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

Ans.

$\begin{array}{l} Radius of circle = 100 cm \\ Length of arc = 22 cm \\ Angle (θ) subtended at the centre by an arc \\ = \frac{Arc}{Radius} radians \\ θ = \frac{22}{100} radians \\ θ = \frac{22}{100} \times \frac{180 °}{π} [∵ 1 radian = \frac{180 °}{π}] \\ θ = \frac{22}{100} \times \frac{180 ° \times 7}{22} \\ = 1 .8 ° \times 7 \\ = 12 .6 ° \\ = 12 ° + 0 .6 \times 60 ‘ \end{array}$ $\begin{array}{l} = 12 ° + 36 ‘ \\ = 12 ° 36 ‘ \\ Thus, angle subtended by an arc of 22 cm at the centre is 12°36′ . \end{array}$

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.

$\begin{array}{l} Diameter of a circle = 40 cm \\ Radius of a circle = \frac{40}{2} cm \\ = 20 cm \\ Length of chord = 20 cm \end{array}$

$\begin{array}{l} Since, Δ OAB is equilateral triangle. \\ So, ∠ AOB = 60 ° \\ = \frac{π}{3} radians \\ Since, angle = \frac{Length of arc}{Radius} \\ \frac{π}{3} Radian = \frac{Length of arc AB}{20 cm} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} Length of arc AB = 20 cm \times \frac{π}{3} \\ = \frac{20 π}{3} cm \end{array}$

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.

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

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.

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

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

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

Ans.

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

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

6. sin 765°
7. cosec (– 1410°)

$8 . \tan \frac{19 π}{3}$

$9 . \sin (- \frac{11 π}{3})$

$10 . \cot (- \frac{15 π}{4})$

Ans.

$\begin{array}{l} 6 . \\ sin (765°) = sin (2 x 36 0 ° + 45°) \\ = sin 45° [∵ \sin (n \times 360 ° + θ) = sinθ] \\ = \frac{1}{\sqrt{2}} \end{array}$ $\begin{array}{l} 7 . \\ \begin{array}{l} cosec (- 1410 °) = - cosec 1410 ° [cosec (- θ) = - cosecθ] \\ = - cosec (4 \times 360 ° - 30 °) \\ = - (- cosec 30 °) [cosec (n \times 360 ° - θ) = - cosecθ] \\ = 2 \end{array} \end{array}$ $\begin{array}{l} 8 . \\ \tan \frac{19 π}{3} = \tan (6 π + \frac{π}{3}) \end{array}$ $\begin{array}{l} \tan \frac{19 π}{3} = \tan \frac{π}{3} [∵ \tan (2 nπ + θ) = tanθ] \\ = \sqrt{3} \end{array}$ $\begin{array}{l} 9 . \\ \begin{array}{l} \sin (- \frac{11 π}{3}) = - \sin (\frac{11 π}{3}) [∵ \sin (- x) = - \sin x] \\ = - \sin (4 π - \frac{π}{3}) \\ = - (- \sin \frac{π}{3}) [∵ \sin (2 nπ - x) = - \sin x] \\ \sin (- \frac{11 π}{3}) = \frac{\sqrt{3}}{2} \end{array} \end{array}$ $\begin{array}{l} 10 . \\ \begin{array}{l} \cot (- \frac{15 π}{4}) = - \cot (\frac{15 π}{4}) [∵ \cot (- x) = - \cot x] \\ = - \cot (4 π - \frac{π}{4}) \\ = - (- \cot \frac{π}{4}) [∵ \cot (2 nπ - x) = - \cot x] \\ = 1 \end{array} \end{array}$

Q.10

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

Ans.

$\begin{array}{l} L . H . S . : \\ \sin^{2} \frac{π}{6} + \cos^{2} \frac{π}{3} - \tan^{2} \frac{π}{4} = {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} - {(1)}^{2} \\ = \frac{1}{4} + \frac{1}{4} - 1 \\ = \frac{1 + 1 - 4}{4} \\ = - \frac{2}{4} \\ = - \frac{1}{2} \\ = R . H . S . Hence proved. \end{array}$

Q.11

$\begin{array}{l} Prove that : \\ 2 \sin^{2} (\frac{π}{6}) + {cosec}^{2} (\frac{7 π}{6}) \cos^{2} (\frac{π}{3}) = \frac{3}{2} \end{array}$

Ans.

$\begin{array}{l} L . H . S . : \\ 2 \sin^{2} \frac{π}{6} + {cosec}^{2} \frac{7 π}{6} \cos^{2} \frac{π}{3} = 2 {(\frac{1}{2})}^{2} + {cosec}^{2} (π + \frac{π}{6}) \times {(\frac{1}{2})}^{2} \\ = 2 \times \frac{1}{4} + \frac{1}{4} \times {cosec}^{2} (\frac{π}{6}) \end{array}$ $\begin{array}{l} = \frac{1}{2} + \frac{1}{4} \times {(2)}^{2} \\ = \frac{1}{2} + 1 \\ = \frac{3}{2} = R . H . S . Hence proved. \end{array}$

Q.12

$\begin{array}{l} Prove that : \\ \cot^{2} \frac{π}{6} + cosec \frac{5 π}{3} + 3 \tan^{2} \frac{π}{6} = 6 \end{array}$

Ans.

$\begin{array}{l} L . H . S : \\ \cot^{2} \frac{π}{6} + cosec \frac{5 π}{6} + 3 \tan^{2} \frac{π}{6} = {(\sqrt{3})}^{2} + cosec (π - \frac{π}{6}) + 3 {(\frac{1}{\sqrt{3}})}^{2} \\ = 3 + cosec (\frac{π}{6}) + 3 \times \frac{1}{3} \\ [∵ cosec (π - x) = cosec x] \\ = 3 + 2 + 1 \\ = 6 = R . H . S . Hence proved. \end{array}$

Q.13

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

Ans.

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

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

Ans.

$\begin{array}{l} (i) \sin 75 ° = \sin (45 ° + 30 °) \\ = \sin 45 ° \cos 30 ° + \sin 30 ° \cos 45 ° \\ [∵ \sin (x + y) = \sin x \cos y + \sin y \cos x] \\ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{\sqrt{2}} \\ = \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} \\ = \frac{\sqrt{3} + 1}{2 \sqrt{2}} \\ (ii) \tan 15 ° = \tan (45 ° - 30 °) \\ = \frac{\tan 45 ° - \tan 30 °}{1 + \tan 45 ° \tan 30 °} \\ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \times \frac{1}{\sqrt{3}}} \end{array}$ $\begin{array}{l} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \\ = \frac{{(\sqrt{3} - 1)}^{2}}{3 - 1} \\ = \frac{3 - 2 \sqrt{3} + 1}{2} \\ \tan 15 ° = \frac{4 - 2 \sqrt{3}}{2} \\ = 2 - \sqrt{3} \end{array}$

Q.15

$\cos (\frac{π}{4} - x) \cos (\frac{π}{4} - y) - \sin (\frac{π}{4} - x) \sin (\frac{π}{4} - y) = \sin (x + y)$

Ans.

$\begin{array}{l} Since, cosA cosB - sinA sinB = \cos (A + B) \\ L . H . S . = \\ \cos (\frac{π}{4} - x) \cos (\frac{π}{4} - y) - \sin (\frac{π}{4} - x) \sin (\frac{π}{4} - y) \\ = \cos {(\frac{π}{4} - x) + (\frac{π}{4} - y)} \\ = \cos (\frac{π}{2} - x - y) \\ = \cos {\frac{π}{2} - (x + y)} \\ = \sin (x + y) [∵ \cos (\frac{π}{2} - θ) = \sin θ] \\ = R . H . S . Hence proved. \end{array}$

Q.16

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

Ans.

$\begin{array}{l} L . H . S : \\ \frac{\tan (\frac{π}{4} + x)}{\tan (\frac{π}{4} - x)} = \frac{(\frac{\tan \frac{π}{4} + \tan x}{1 - \tan \frac{π}{4} \tan x})}{(\frac{\tan \frac{π}{4} - \tan x}{1 + \tan \frac{π}{4} \tan x})} [\begin{array}{l} ∵ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \\ \tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \end{array}] \\ = \frac{(\frac{1 + \tan x}{1 - 1 . \tan x})}{(\frac{1 - \tan x}{1 + 1 . \tan x})} \\ = (\frac{1 + \tan x}{1 - \tan x}) \times (\frac{1 + \tan x}{1 - \tan x}) \\ = {(\frac{1 + \tan x}{1 - \tan x})}^{2} \\ = R . H . S . Hence proved. \end{array}$

Q.17

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

Ans.

$\begin{array}{l} L . H . S . : \\ \frac{\cos (π + x) \cos (- x)}{\sin (π - x) \cos (\frac{π}{2} + x)} = \frac{- \cos x \cos x}{\sin x \times - \sin x} [\begin{array}{l} ∵ \cos (π + x) = - cosx \\ \cos (- x) = cosx \\ \sin (π - x) = sinx \\ \cos (\frac{π}{2} + x) = - sinx \end{array}] \\ = \cot^{2} x \\ = R . H . S . Hence proved. \end{array}$

Q.18

$\cos (\frac{3 π}{2} + x) \cos (2 π + x) [\cot (\frac{3 π}{2} - x) + \cot (2 π + x)] = 1$

Ans.

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

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

Ans.

$\begin{array}{l} L . H . S . : sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x \\ = cos (n + 1) x cos (n + 2) x + sin (n + 1) x sin (n + 2) x \\ = cos {(n + 1) x - (n + 2) x} \\ [∵ cosA cosB - sinA sinB = \cos (A + B)] \\ = cos (nx + x - nx - 2 x) \\ = cos (- x) [∵ \cos (- A) = cosA] \\ = \cos x = R . H . S . \end{array}$

Q.20

$\cos (\frac{3 π}{4} + x) - \cos (\frac{3 π}{4} - x) = - \sqrt{2} \sin x$

Ans.

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

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

Ans.

$\begin{array}{l} L . H . S . = {sin}^{2} 6x - {sin}^{2} 4x \\ = (\sin 6 x + \sin 4 x) (\sin 6 x - \sin 4 x) [∵ a^{2} - b^{2} = (a - b) (a + b)] \\ = {2 \sin (\frac{6 x + 4 x}{2}) \cos (\frac{6 x - 4 x}{2})} {2 \cos (\frac{6 x + 4 x}{2}) \sin (\frac{6 x - 4 x}{2})} \\ = {2 \sin 5 x \cos x} {2 \cos 5 x \sin x} \\ = (2 \sin x \cos x) (2 \sin 5 x \cos 5 x) \\ = \sin 2 x . \sin 10 x [∵ \sin 2 x = 2 \sin x \cos x] \\ = R . H . S . Hence proved. \end{array}$

Q.22 Prove that:
cos²2x – cos²6x = sin 4x sin 8x

Ans.

$\begin{array}{l} L . H . S . = \cos^{2} 2x - \cos^{2} 6x \\ = (\cos 2 x + \cos 6 x) (\cos 2 x - \cos 6 x) \end{array}$ $\begin{array}{l} = {2 \cos (\frac{2 x + 6 x}{2}) \cos (\frac{2 x - 6 x}{2})} {- 2 \sin (\frac{2 x - 6 x}{2}) \sin (\frac{2 x + 6 x}{2})} \\ = {2 \cos 4 x \cos (- 2 x)} {- 2 \sin (- 2 x) \sin 4 x} \\ = (2 \sin 2 x \cos 2 x) (2 \sin 4 x \cos 4 x) \\ = \sin 4 x . \sin 8 x [∵ \sin 2 x = 2 \sin x \cos x] \\ = R . H . S . Hence proved. \end{array}$

Q.23 Prove that:
sin 2x + 2 sin 4x + sin 6x = 4 cos²x sin 4x

Ans.

$\begin{array}{l} sin 2x + 2 sin 4x + sin 6x = 2 sin 4x + sin 6x + sin 2x \\ = 2 sin 4x + 2 \sin (\frac{6 x + 2 x}{2}) \cos (\frac{6 x - 2 x}{2}) \\ = 2 sin 4x + 2 \sin 4 x \cos 2 x \\ = 2 sin 4x (1 + \cos 2 x) \\ = 2 sin 4x (2 \cos^{2} x) \\ = 4 \cos^{2} x sin 4x \\ = R . H . S . \end{array}$

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

Ans.

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

Q.25

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

Ans.

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

Q.26

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

Ans.

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

Q.27

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

Ans.

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

Q.28

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

Ans.

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

Q.29

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

Ans.

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

Q.30

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

Ans.

$\begin{array}{l} L . H . S . : \\ \frac{\cos 4 x + \cos 3 x + \cos 2 x}{\sin 4 x + \sin 3 x + \sin 2 x} = \frac{\cos 4 x + \cos 2 x + \cos 3 x}{\sin 4 x + \sin 2 x + \sin 3 x} \\ = \frac{2 \cos (\frac{4 x + 2 x}{2}) \cos (\frac{4 x - 2 x}{2}) + \cos 3 x}{2 \sin (\frac{4 x + 2 x}{2}) \cos (\frac{4 x - 2 x}{2}) + \sin 3 x} \\ = \frac{2 \cos 3 x \cos x + \cos 3 x}{2 \sin 3 x \cos x + \sin 3 x} \end{array}$ $\begin{array}{l} = \frac{\cos 3 x (2 \cos x + 1)}{\sin 3 x (2 \cos x + 1)} \\ \frac{\cos 4 x + \cos 3 x + \cos 2 x}{\sin 4 x + \sin 3 x + \sin 2 x} = \cot 3 x = R . H . S . Hence proved. \end{array}$

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

Ans.

$\begin{array}{l} ∵ \cot 3 x = \cot (2 x + x) \\ = \frac{\cot 2 x \cot x - 1}{\cot 2 x + \cot x} \\ \cot 3 x (\cot 2 x + \cot x) = \cot 2 x \cot x - 1 \\ \cot 3 x \cot 2 x + \cot 3 x \cot x = \cot 2 x \cot x - 1 \\ 1 = \cot x \cot 2 x - \cot 2 x \cot 3 x - \cot 3 x \cot x \\ or \cot x \cot 2 x - \cot 2 x \cot 3 x - \cot 3 x \cot x = 1 . Hence proved. \end{array}$

Q.32

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

Ans.

$\begin{array}{l} L . H . S : \\ \tan 4 x = \tan 2 (2 x) \\ = \frac{2 \tan 2 x}{1 - \tan^{2} 2 x} \end{array}$ $\begin{array}{l} = \frac{2 (\frac{2 \tan x}{1 - \tan^{2} x})}{1 - {(\frac{2 \tan x}{1 - \tan^{2} x})}^{2}} \\ = \frac{(\frac{4 \tan x}{1 - \tan^{2} x})}{{\frac{{(1 - \tan^{2} x)}^{2} - 4 \tan^{2} x}{{(1 - \tan^{2} x)}^{2}}}} \\ = \frac{4 \tan x}{1 - \tan^{2} x} \times \frac{{(1 - \tan^{2} x)}^{2}}{1 - 2 \tan^{2} x + \tan^{4} x - 4 \tan^{2} x} \\ = \frac{4 \tan x (1 - \tan^{2} x)}{1 - 6 \tan^{2} x + \tan^{4} x} = R . H . S . \\ Hence proved. \end{array}$

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

Ans.

$\begin{array}{l} L . H . S . : \cos 4 x = \cos 2 (2 x) \\ = 1 - 2 \sin^{2} (2 x) [∵ \cos 2 x = 1 - \sin^{2} x] \\ = 1 - 2 {(2 \sin x \cos x)}^{2} [∵ \sin 2 x = 2 \sin x \cos x] \\ = 1 - 2 (4 \sin^{2} x \cos^{2} x) \\ = 1 - 8 \sin^{2} x \cos^{2} x \\ = R . H . S . \end{array}$

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

Ans.

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

Q.35

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

Ans.

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

Q.36

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

Ans.

$\begin{array}{l} Since, secx = 2 \Rightarrow \sec \frac{π}{3} = 2 and sec \frac{5 π}{3} = \sec (2 π - \frac{π}{3}) = \sec \frac{π}{3} = 2 \\ Therefore, principal solutions are x = \frac{π}{3} and \frac{5 π}{3} . \\ G eneral solution: \\ secx = 2 \\ = \sec \frac{π}{3} \\ = \sec (2 nπ \pm \frac{π}{3}) \\ ∴ x = 2 nπ \pm \frac{π}{3}, where n \in Z . \end{array}$

Q.37

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

Ans.

$\begin{array}{l} Since, cot x = - \sqrt{3} \\ \Rightarrow cot x = - \cot \frac{π}{6} \\ = \cot (π - \frac{π}{6}) \\ = \cot (\frac{5 π}{6}) \\ and \\ cot x = \cot (2 π - \frac{π}{6}) \\ = \cot (\frac{11 π}{6}) \\ Therefore, the principal solution are x = \frac{5 π}{6} and \frac{11 π}{6} . \\ General solution: \\ cotx = - \sqrt{3} \\ = \cot (\frac{5 π}{6}) \\ = \cot (nπ + \frac{5 π}{6}) \\ \Rightarrow x = nπ + \frac{5 π}{6}, where n \in Z . \end{array}$

Q.38

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

Ans.

$Since, cosec x = cosec (\frac{π}{6})$ $\begin{array}{l} ∴ cosec x = - 2 \\ = - cosec (\frac{π}{6}) \\ = cosec (π + \frac{π}{6}) \\ = cosec (\frac{7 π}{6}) \\ and \\ cosec x = cosec (2 π - \frac{π}{6}) \\ = cosec \frac{11 π}{6} \\ Therefore, the principal solutions are \frac{7 π}{6}, \frac{11 π}{6} . \\ General solution: \\ cosec x = cosec (\frac{7 π}{6}) \\ = cosec (nπ + {(- 1)}^{n} \frac{7 π}{6}) \\ \Rightarrow x = nπ + {(- 1)}^{n} \frac{7 π}{6}, n \in Z . \end{array}$

Q.39

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

Ans.

$\begin{array}{l} We have, \\ cos 4x = cos 2x \end{array}$ $\begin{array}{l} = \cos (2 nπ \pm 2 x) [\begin{array}{l} The general solution of cos θ = \cos α \\ \Rightarrow θ = 2 nπ \pm α \end{array}] \\ \Rightarrow 4 x = 2 nπ \pm 2 x \\ For (+) ve sign: \\ 4 x = 2 nπ + 2 x \\ 2 x = 2 nπ \Rightarrow x = nπ \\ For (-) ve sign: \\ 4 x = 2 nπ - 2 x \\ 6 x = 2 nπ \Rightarrow x = \frac{1}{3} nπ \\ The general solutions are x = \frac{1}{3} nπ and nπ, where n \in Z . \end{array}$

Q.40

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

Ans.

$\begin{array}{l} We have, \cos 3 x + \cos x - \cos 2 x = 0 \\ 2 \cos (\frac{3 x + x}{2}) \cos (\frac{3 x - x}{2}) - \cos 2 x = 0 \\ 2 \cos 2 x \cos x - \cos 2 x = 0 \\ \cos 2 x (2 \cos x - 1) = 0 \\ \Rightarrow \cos 2 x = 0 \Rightarrow 2 x = (2 n + 1) \frac{π}{2} \\ \Rightarrow x = (2 n + 1) \frac{π}{4}, n \in Z \end{array}$ $\begin{array}{l} or 2 cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2} = \cos \frac{π}{3} \\ \Rightarrow x = 2 nπ \pm \frac{π}{3}, n \in Z . \end{array}$

Q.41

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

Ans.

$\begin{array}{l} We have, \sin 2 x + \cos x = 0 \\ \Rightarrow 2 \sin x \cos x + \cos x = 0 \\ \Rightarrow \cos x (2 \sin x + 1) = 0 \\ Either \cos x = 0 \Rightarrow x = (2 n + 1) \frac{π}{2}, n \in Z . \\ or (2 \sin x + 1) = 0 \Rightarrow \sin x = - \frac{1}{2} \\ \Rightarrow \sin x = \sin \frac{7 π}{2} \\ \Rightarrow x = nπ + {(- 1)}^{n} \frac{7 π}{2}, n \in Z . \\ Thus, the general solution of given equation is: \\ nπ + {(- 1)}^{n} \frac{7 π}{2} or (2 n + 1) \frac{π}{2}, n \in Z . \end{array}$

Q.42

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

Ans.

$\begin{array}{l} We have, \sec^{2} 2 x = 1 - \tan 2 x \\ \Rightarrow 1 + \tan^{2} 2 x = 1 - \tan 2 x \end{array}$ $\begin{array}{l} \Rightarrow \tan^{2} 2 x + \tan 2 x = 0 \\ \Rightarrow \tan 2 x (\tan 2 x + 1) = 0 \\ Either tan 2x = 0 \Rightarrow 2 x = nπ \\ \Rightarrow x = \frac{nπ}{2}, where n \in Z . \\ or \tan 2 x + 1 = 0 \Rightarrow \tan 2 x = - 1 \\ \Rightarrow \tan 2 x = - \tan \frac{π}{4} \\ = \tan (π - \frac{π}{4}) \\ \Rightarrow \tan 2 x = \tan \frac{3 π}{4} \\ \Rightarrow 2 x = nπ + \frac{3 π}{4} \\ \Rightarrow x = \frac{nπ}{2} + \frac{3 π}{8}, where n \in Z . \\ The general solution of the given equation is: \\ x = \frac{nπ}{2}, \frac{nπ}{2} + \frac{3 π}{8}, where n \in Z . \end{array}$

Q.43

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

Ans.

$\begin{array}{l} We have, \sin x + \sin 3 x + \sin 5 x = 0 \\ \Rightarrow \sin 5 x + \sin x + \sin 3 x = 0 \\ \Rightarrow 2 \sin (\frac{5 x + x}{2}) \cos (\frac{5 x - x}{2}) + \sin 3 x = 0 \end{array}$ $\begin{array}{l} \Rightarrow 2 \sin 3 x \cos 2 x + \sin 3 x = 0 \\ \Rightarrow \sin 3 x (2 \cos 2 x + 1) = 0 \\ Either sin3x = 0 \Rightarrow 3 x = nπ \\ \Rightarrow x = \frac{nπ}{3}, n \in Z . \\ or 2 cos 2x + 1 = 0 \Rightarrow \cos 2 x = - \frac{1}{2} \\ \Rightarrow 2 x = 2 nπ \pm \frac{2 π}{3} \\ \Rightarrow x = nπ \pm \frac{π}{3}, n \in Z . \\ Therefore, the general solution of the given equations is: \\ x = \frac{nπ}{3} or nπ \pm \frac{π}{3}, n \in Z . \end{array}$

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.

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

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

Q.45

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

Q.46

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

Ans.

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

Q.47

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

Ans.

$\begin{array}{l} We have, \\ \frac{b - c}{a} = \frac{k \sin B - k \sin C}{k \sin A} [∵ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k (let)] \\ = \frac{\sin B - \sin A}{\sin A} \\ = \frac{2 \cos (\frac{B + C}{2}) \sin (\frac{B - C}{2})}{2 \sin \frac{A}{2} \cos \frac{A}{2}} \\ = \frac{\cos (\frac{π}{2} - \frac{A}{2}) \sin (\frac{B - C}{2})}{\sin \frac{A}{2} \cos \frac{A}{2}} \\ = \frac{\sin (\frac{A}{2}) \sin (\frac{B - C}{2})}{\sin \frac{A}{2} \cos \frac{A}{2}} \\ = \frac{\sin (\frac{B - C}{2})}{\cos \frac{A}{2}} \end{array}$ $\begin{array}{l} \sin (\frac{B - C}{2}) = \frac{b - c}{a} \cos \frac{A}{2} \\ Therefore, it is proved. \end{array}$

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

Ans.

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

Q.49

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

Ans.

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

Q.50

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

Ans.

$\begin{array}{l} R . H . S . = \frac{b^{2} - c^{2}}{a^{2}} \\ = \frac{k^{2} \sin^{2} B - k^{2} \sin^{2} C}{k^{2} \sin^{2} A} [∵ \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC} = k (let)] \\ = \frac{\sin^{2} B - \sin^{2} C}{\sin^{2} A} \\ = \frac{\sin (B + C) \sin (B - C)}{\sin^{2} A} [∵ \sin^{2} x - \sin^{2} y = \sin (x + y) \sin (x - y)] \end{array}$ $\begin{array}{l} = \frac{\sin (π - A) \sin (B - C)}{\sin^{2} A} [∵ B + C = π - A] \\ = \frac{\sin A \sin (B - C)}{\sin^{2} A} [∵ \sin (π - A) = sinA] \\ = \frac{\sin (B - C)}{\sin {π - (B + C)}} [∵ A = π - (B + C)] \\ = \frac{\sin (B - C)}{\sin (B + C)} [∵ \sin (π - θ) = sinθ] \\ = L . H . S . \\ Thus, it is proved. \end{array}$

Q.51

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

Ans.

$\begin{array}{l} We are given : (b + c) \cos \frac{B + C}{2} = a \cos \frac{B - C}{2} \\ or \frac{(b + c)}{a} = \frac{\cos \frac{B - C}{2}}{\cos \frac{B + C}{2}} \\ L . H . S . = \frac{(b + c)}{a} \\ = \frac{k \sin B + k \sin C}{k \sin A} [∵ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k (let)] \\ = \frac{\sin B + \sin C}{\sin A} \\ = \frac{2 \sin (\frac{B + C}{2}) \cos (\frac{B - C}{2})}{2 \sin \frac{A}{2} \cos \frac{A}{2}} [\begin{array}{l} \sin x + \sin y \\ = 2 \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2}) \end{array}] \\ = \frac{\sin (\frac{π}{2} - \frac{A}{2}) \cos (\frac{B - C}{2})}{\sin \frac{A}{2} \cos \frac{A}{2}} \\ = \frac{\cos \frac{A}{2} \cos (\frac{B - C}{2})}{\sin (\frac{π}{2} - \frac{B + C}{2}) \cos (\frac{A}{2})} [∵ \sin (\frac{π}{2} - θ) = cosθ] \end{array}$ $\begin{array}{l} = \frac{\cos (\frac{B - C}{2})}{\cos (\frac{B + C}{2})} = R . H . S . \\ Thus, (b + c) \cos \frac{B + C}{2} = a \cos \frac{B - C}{2} is proved. \end{array}$

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

Ans.

$\begin{array}{l} L . H . S . = a cos A + b cos B + c cos C \\ = k \sin A \cos A + k \sin B \cos B + k \sin C \cos C \\ [∵ \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC} = k (let)] \\ = \frac{k}{2} (2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C) \\ = \frac{k}{2} (\sin 2 A + \sin 2 B + \sin 2 C) \\ = \frac{k}{2} (2 \sin \frac{2 A + 2 B}{2} . \cos \frac{2 A - 2 B}{2} + \sin 2 C) \\ = \frac{k}{2} {2 \sin (A + B) . \cos (A - B) + \sin 2 C} \\ = \frac{k}{2} {2 \sin (π - C) . \cos (A - B) + 2 \sin C \cos C} \\ [∵ A + B + C = π] \\ = \frac{k}{2} {2 \sin C . \cos (A - B) + 2 \sin C \cos (π - A - B)} \\ [∵ \sin (π - θ) = sinθ] \end{array}$ $\begin{array}{l} = k \sin C {\cos (A - B) - \cos (A + B)} \\ = k \sin C [2 \sin (\frac{A - B + A + B}{2}) \sin {\frac{A + B - (A - B)}{2}}] \\ = 2 k \sin C . \sin A . \sin B \\ = 2 (k \sin A) . \sin B . \sin C \\ = 2 a \sin B . \sin C [\frac{\sin A}{a} = k (let)] \\ = R . H . S . \\ Thus, it is proved. \end{array}$

Q.53

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

Ans.

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

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

Ans.

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

Q.55

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

Ans.

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

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.

$\begin{array}{l} ∠ APR = 90 ° - 60 ° \\ = 30 ° \end{array}$ $\begin{array}{l} \Rightarrow ∠ APQ = 30 ° \\ ∠ PAQ = 60 ° - 15 ° \\ = 45 ° \\ ∠ PQA = 180 ° - ∠ PAQ - ∠ APQ \\ = 180 ° - 45 ° - 30 ° \\ = 105 ° \\ Applying sin rule in ΔPQA, we get \\ \frac{\sin 45 °}{PQ} = \frac{\sin 30 °}{AQ} = \frac{\sin 105 °}{AP} \\ \Rightarrow \frac{\sin 45 °}{PQ} = \frac{\sin 30 °}{35} \\ \Rightarrow \frac{(\frac{1}{\sqrt{2}})}{PQ} = \frac{(\frac{1}{2})}{35} \\ \Rightarrow \frac{35}{\sqrt{2}} = \frac{PQ}{2} \\ \Rightarrow PQ = 35 \sqrt{2} \\ Thus, the height of the tree is 35 \sqrt{2} m . \end{array}$

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°

$\begin{array}{l} By using cosine formula in ΔOPQ, we get \\ {PQ}^{2} = {OP}^{2} + {OQ}^{2} - 2 \times OP \times OQ \cos 60 ° \\ = {(72)}^{2} + {(96)}^{2} - 2 \times 72 \times 96 \times \frac{1}{2} \\ = 5184 + 9216 - 6912 \\ PQ = \sqrt{7488} \\ = 86 .53 km \end{array}$

Q.58

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

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°.

$\begin{array}{l} Applying cosine formula in ΔABC, we get \\ \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 ab} \\ \cos 45 ° = \frac{{(300)}^{2} + {(250)}^{2} - c^{2}}{2 (300) (250)} \\ \frac{1}{\sqrt{2}} = \frac{90000 + 62500 - c^{2}}{150000} \end{array}$ $\begin{array}{l} \frac{150000}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 152500 - {AB}^{2} \\ \frac{150000 \sqrt{2}}{2} = 152500 - {AB}^{2} \\ \frac{150000 \times 1 .414}{2} = 152500 - {AB}^{2} \\ 106050 = 152500 - {AB}^{2} \\ AB = \sqrt{152500 - 106050} \\ = \sqrt{46450} \\ AB = 215 .5 m \\ Thus, distance between two trees is 215.5m. \end{array}$

Q.59 Prove that:

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

Ans.

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

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

Ans.

$\begin{array}{l} L . H . S . = (\sin 3 x + \sin x) sinx + (\cos 3 x - \cos x) \cos x \\ = \sin 3 x \sin x + \sin^{2} x + \cos 3 x \cos x - \cos^{2} x \\ = \cos 3 x \cos x + \sin 3 x \sin x - \cos^{2} x + \sin^{2} x \\ = \cos (3 x - x) - (\cos^{2} x - \sin^{2} x) \\ [∵ \cos (A - B) = \cos A \cos B + \sin A \cos B] \\ = \cos 2 x - \cos 2 x [∵ \cos^{2} x - \sin^{2} x = \cos 2 x] \\ = 0 \\ = R . H . S . Hence proved. \end{array}$

Q.61 Prove that:

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

Ans.

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

Q.62 Prove that:

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

Ans.

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

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

Ans.

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

Q.64

Prove that:

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

Ans.

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

Q.65

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

Ans.

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

Q.66

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

Ans.

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

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

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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.

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)

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.

NCERT Solutions Class 11 Maths Chapter 3

NCERT Solutions Class 11 Maths Chapter 3 – Trigonometric Functions

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

NCERT Solutions Class 11 Maths Chapter 3: Exercise & Solutions

NCERT Exemplar Class 11 Maths

Key Features of NCERT Solutions Class 11 Maths Chapter 3

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NCERT Solutions Class 11 Maths Chapter 3

NCERT Solutions Class 11 Maths Chapter 3 – Trigonometric Functions

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

NCERT Solutions Class 11 Maths Chapter 3: Exercise & Solutions

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Key Features of NCERT Solutions Class 11 Maths Chapter 3

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