# NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1

Many students are intimidated by the subject of Mathematics. The concepts, formulas, theorems, and calculations of the subject are challenging for them. Students should have a thorough understanding of the topics in Mathematics since it is a purely conceptual subject. Chapter 8 Class 10 Mathematics is Trigonometry. Students may find it difficult to understand the concepts and calculations in this chapter at first. However, the chapter’s concepts eventually become easier for them with meticulous practise and proper guidance. The curriculum of Chapter-8, Class 10 Mathematics includes the following topics – The Introduction of the Chapter, Trigonometric Ratios, Trigonometric Ratios of Some Specific Angles, Trigonometric Ratios of Complementary Angles, Trigonometric Identities and summary. This is one of the most challenging chapters for the students of Class 10, as it is very difficult for them to grasp the concepts of the chapter. To ensure that students perform effectively in their board examinations, Extramarks provides them with all the essential resources, such as NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 to make their learning experience comprehensive and effortless.

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## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Ex 8.1) Exercise 8.1

The Central Board of Secondary Education is a major board of education under the control of the Government of India. In India, the Central Board of Secondary Education controls, manages, and oversees the school education system and matters related to the academic curriculum. All the schools that are associated with CBSE, especially Classes 9 to 12, follow the NCERT curriculum. Each year, between February and May, CBSE conducts board examinations for Classes 10 and 12. The NCERT curriculum is followed by all schools affiliated with the CBSE board. In 1961, the Indian Government established NCERT as an independent organisation to assist the Central and State Governments in improving the quality of education in schools. In the areas of education, NCERT is primarily responsible for controlling and coordinating research education, publishing NCERT textbooks, and multimedia digital materials, preparing supplementary material, and much more to help students achieve a successful academic career.Teachers are also trained in innovative educational practises through the organisation’s training programs. Additionally, NCERT conducts bilateral cultural exchanges with other countries in the field of school education, as well as research, development, training, extension, publication, and dissemination. The NCERT also collaborates and interacts with international organisations, visits foreign delegations, and provides various training facilities  for educators in developing countries.

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## Class 10 Maths Chapter 8 Exercise 8.1: Topics Covered

The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 encompass basic concepts of the specific functions of angles and their application to calculations. It is one of the most used concepts in the fields of Astronomy and Geography. The concept is further applicable in the fields of, Satellite Navigation, Chemistry Number Theory, Electronics, Medical Imaging, Electrical Engineering, Architecture, Mechanical Engineering, Civil Engineering, Oceanography, Seismology, and so on and so forth. Students can refer to the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 to have a deeper understanding of the topics of the chapter. Chapter 8 Class 10 Mathematics includes the following topics- Introduction of the Chapter, and Trigonometric Ratios. Even though there are a limited number of topics in the exercise, students may find it difficult to comprehend the concepts. To assist students with the preparation for board and competitive examinations, NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 are an essential resource. Furthermore, students who want to pursue Mathematics for further studies or want to perform well in any competitive examination should review the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 very keenly.

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Trigonometry is the study of the relationship between the sides and angles of triangles, as well as the relevant functions of angles. Students can refer to NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 for a better understanding of Chapter 8 Trigonometry. In order to prepare students effectively for board examinations, Extramarks offers them a variety of learning modules. Extramarks is a paradigm of the benefits of e-learning. K12 Study Material provided by Extramarks helps students score well in examinations by providing complete and convenient study materials to them. Through Extramarks, students are able to self-assess their performance and track their academic progress through In-depth Performance Reports. In Live Doubt-Solving classes, mentors of Extramarks provide students with the facility of attending interactive sessions. As a result, they can perform well in the examinations by resolving their queries right away. To clear up their doubts, students can also refer to the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. Extramarks gives students access to chapter-based worksheets, unlimited practise questions, interactive activities, and more, so they can become proficient in all subjects. A Visual Learning Journey is also provided by the Learning App so that students can quickly grasp and engage with their subject’s curriculum. Educators at Extramarks provide students with the best possible guidance in their respective fields.

### What are Trigonometric Identities?

Trigonometric Identities are a set of equalities involving trigonometry functions that hold true for all values of variables arranged in an equation. Each trigonometric identity is based on trigonometric ratios. The trigonometric ratios are cosine, tangent, sine, secant, cosecant, and cotangent. The sides of the right triangle are used to define these trigonometric ratios. Students can use NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 to develop the foundational skills they need to solve questions accurately and efficiently in their examinations. A comprehensive course is provided by Extramarks to learners to ensure their success in examinations. The Extramarks Learning Appis available for convenient download. This application is easy to use and a great online learning tool. Students can refer to the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 provided by Extramarks for a better understanding of the chapter.

### What Are Trigonometry Ratios?

In a right-angled triangle, trigonometric ratios represent the values of all trigonometric functions. The trigonometric ratios of a right-angled triangle are the ratios of its sides to its acute angles. The adjacent side is the base of the right triangle, and the opposite side is perpendicular. With the help of the Extramarks website, students can easily access NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 at any time and from any place. A simple language has been used to compile these solutions. A comprehensive understanding of the basic concepts of a curriculum is essential for students to do well in a subject. In order to make learning easier for students, Extramarks provides them with NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. In order to master the basics of various mathematical themes, the student should familiarise themselves with the NCERT textbook. Comprehensive step-by-step explanations and detailed answers are provided on the Extramarks website to help students better understand the concepts.Students can rely on these solutions since they have been compiled by subject matter experts of Extramarks. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 are provided on Extramarks, a website that provides reliable tools for preparing for board examinations.

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## Access NCERT Solutions for Class -10 Maths Chapter 8 – Introduction to Trigonometry

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## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1

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### Exercise 8.1 Class 10 Solutions Chapter 8 Trigonometry- Free PDF Download

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### Class 10th Maths Chapter 8 Exercise 8.1

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### NCERT Solutions for Class 10 Maths Chapter 8 All Other Exercises

Students can improve their knowledge of Chapter 8-Trigonometry by practising NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1. Apart from these solutions, the other exercises of the chapter help students to understand the fundamentals of trigonometry and also assist them in applying those concepts practically. The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 and the other exercises are designed for students to improve their mathematical concepts and calculation abilities efficiently. By going through these exercises, students can practise the concepts and functions of the chapter, which are very essential to succeed in any in-school, board, or competitive examinations. Practising the NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 and the other exercises is the most important step that students should take in preparation for the chapter  their examinations.

Q.1

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{\hspace{0.17em}ABC, right-angled at B, AB}=\text{24 cm, BC}=\text{7 cm.}\\ \text{Determine :}\\ \text{(i) sin A, cos A}\\ \text{(ii) sin C, cos C}\end{array}$

Ans

$\begin{array}{l}\text{Using Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ \text{AC\hspace{0.17em}}=\sqrt{{24}^{2}+{7}^{2}}=\sqrt{576+49}=\sqrt{625}=25\text{cm}\end{array}$

$\begin{array}{l}\text{(i)}\\ \text{sin\hspace{0.17em}A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}\\ \mathrm{cos}\text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}\\ \text{(ii)}\\ \text{sin\hspace{0.17em}C}=\frac{\text{side opposite to angle C}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}\\ \mathrm{cos}\text{C}=\frac{\text{side adjacent to angle C}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}\end{array}$

Q.2

$\text{In the following figure},\text{}\mathrm{tanP}-\mathrm{cotR}.$

Ans

$\begin{array}{l}\text{Using Pythagoras theorem in}\mathrm{\Delta }\text{PQR, we get}\\ \text{QR\hspace{0.17em}}=\sqrt{{13}^{2}-{12}^{2}}=\sqrt{169-144}=\sqrt{25}=5\text{cm}\end{array}$

$\begin{array}{l}\mathrm{tanP}=\frac{\text{side opposite to angle P}}{\text{side adjacent to angle P}}=\frac{\text{QR}}{\text{PQ}}=\frac{5}{12}\\ \mathrm{cotR}=\frac{\text{side adjacent to angle R}}{\text{side opposite to angle R}}=\frac{\text{QR}}{\text{PQ}}=\frac{5}{12}\\ \\ \mathrm{tanP}-\mathrm{cotR}=\frac{5}{12}-\frac{5}{12}=0\end{array}$

Q.3

$\text{If sin A}=\frac{3}{4},\text{calculate}\mathrm{cos}\text{A and}\mathrm{tan}\text{A.}$

Ans

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right-angled triangle right angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{sin A}=\frac{3}{4}\\ \text{or}\frac{\text{BC}}{\text{AC}}=\frac{3}{4}\\ \text{Let BC be 3k. Therefore, AC will be 4k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{or (4k)}}^{2}={\text{AB}}^{2}+{\left(3\mathrm{k}\right)}^{2}\\ \text{or}16{\mathrm{k}}^{2}-9{\mathrm{k}}^{2}={\text{AB}}^{2}\\ \text{or \hspace{0.17em}}7{\mathrm{k}}^{2}={\text{AB}}^{2}\\ \text{or \hspace{0.17em} AB}=\sqrt{7}\mathrm{k}\\ \mathrm{cos}\text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{7}\mathrm{k}}{4\mathrm{k}}=\frac{\sqrt{7}}{4}\\ \mathrm{tan}\text{A}=\frac{\text{side opposite to angle A}}{\text{side adjacent to angle A}}=\frac{\text{BC}}{\text{AB}}=\frac{3\mathrm{k}}{\sqrt{7}\mathrm{k}}=\frac{3}{\sqrt{7}}\end{array}$

Q.4

$\text{Given 15 cot A}=\text{8, find sin A and sec A.}$

Ans

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right-angled triangle right angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{15 cot A}=\text{8}\\ \text{or cot A}=\frac{8}{15}\\ \text{or}\frac{\text{AB}}{\text{BC}}=\frac{8}{15}\\ \text{Let AB be 8k. Therefore, BC will be 15k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{or AC}}^{2}={\left(\text{15k}\right)}^{2}+{\left(8\mathrm{k}\right)}^{2}\\ {\text{or AC}}^{2}=225{\mathrm{k}}^{2}+64{\mathrm{k}}^{2}\\ {\text{or \hspace{0.17em}AC}}^{2}=\text{289}{\mathrm{k}}^{2}\\ \text{or \hspace{0.17em}AC}=17\mathrm{k}\\ \mathrm{sin}\text{A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{15\mathrm{k}}{17\mathrm{k}}=\frac{15}{17}\\ \mathrm{sec}\text{A}=\frac{\text{hypotenuse}}{\text{side adjacent to angle A}}=\frac{\text{AC}}{\text{AB}}=\frac{17\mathrm{k}}{8\mathrm{k}}=\frac{17}{8}\end{array}$

Q.5

$\text{Given sec}\mathrm{\theta }=\frac{13}{12}\text{, calculate all other trigonometric ratios.}$

Ans

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right triangle, right-angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{sec}\mathrm{\theta }=\frac{13}{12}\\ \text{or}\frac{\text{AC}}{\text{AB}}=\frac{13}{12}\\ \text{Let AC be 13k. Therefore, AB is 12k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{BC}}^{2}={\text{AC}}^{2}-{\text{AB}}^{2}\\ {\text{or BC}}^{2}={\left(\text{13k}\right)}^{2}-{\left(12\mathrm{k}\right)}^{2}\\ {\text{or BC}}^{2}=169{\mathrm{k}}^{2}-144{\mathrm{k}}^{2}\\ {\text{or \hspace{0.17em}BC}}^{2}=\text{25}{\mathrm{k}}^{2}\\ \text{or \hspace{0.17em}BC}=5\mathrm{k}\\ \mathrm{sin\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{5\mathrm{k}}{13\mathrm{k}}=\frac{5}{13}\\ \mathrm{cos\theta }=\frac{\text{side adjacent to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{12\mathrm{k}}{13\mathrm{k}}=\frac{12}{13}\\ \mathrm{tan\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{side adjacent to angle}\mathrm{\theta }}=\frac{\text{BC}}{\text{AB}}=\frac{5\mathrm{k}}{12\mathrm{k}}=\frac{5}{12}\\ \mathrm{cot\theta }=\frac{\text{side adjacent to angle}\mathrm{\theta }}{\text{side opposite to angle}\mathrm{\theta }}=\frac{\text{AB}}{\text{BC}}=\frac{12\mathrm{k}}{5\mathrm{k}}=\frac{12}{5}\\ \mathrm{cosec}\text{\hspace{0.17em}}\mathrm{\theta }=\frac{\text{hypotenuse}}{\text{side opposite to angle}\mathrm{\theta }}=\frac{\text{AC}}{\text{BC}}=\frac{13\mathrm{k}}{5\mathrm{k}}=\frac{13}{5}\end{array}$

Q.6

$\begin{array}{l}\text{If}\angle \text{A and}\angle \text{B are acute angles such that cos A}=\text{cos B,}\\ \text{then show that}\angle \text{A}=\angle \text{B.}\end{array}$

Ans

$\text{Let us consider a}\mathrm{\Delta }\text{\hspace{0.17em}ABC in which CD}\perp \text{AB.}$

$\begin{array}{l}\text{Given that,}\\ \text{cos A}=\text{cos B}\\ \text{or}\frac{\text{AD}}{\text{AC}}=\frac{\text{BD}}{\text{BC}}\\ \text{or}\frac{\text{AD}}{\text{BD}}=\frac{\text{AC}}{\text{BC}}\\ \text{Let}\frac{\text{AD}}{\text{BD}}=\frac{\text{AC}}{\text{BC}}=\mathrm{k}\text{}\\ \text{or AD}=\mathrm{k}\text{BD}...\text{(1)}\\ \text{and}\\ \text{AC}=\mathrm{k}\text{BC}...\text{(2)}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{\hspace{0.17em}CAD and}\mathrm{\Delta }\text{\hspace{0.17em}CBD, we get}\\ {\text{CD}}^{2}={\text{AC}}^{2}-{\text{AD}}^{2}\text{}...\text{(3)}\\ \text{and}\\ {\text{CD}}^{2}={\text{BC}}^{2}-{\text{BD}}^{2}\text{}...\text{(4)}\\ \text{From equations (3) and (4), we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}}^{2}-{\text{AD}}^{2}={\text{BC}}^{2}-{\text{BD}}^{2}\\ \text{(}\mathrm{k}{\text{BC)}}^{2}-{\left(\mathrm{k}\text{BD}\right)}^{2}={\text{BC}}^{2}-{\text{BD}}^{2}\\ \text{or}{\mathrm{k}}^{2}{\text{(BC}}^{2}-{\text{BD}}^{2}\right)={\text{BC}}^{2}-{\text{BD}}^{2}\\ \text{or}{\mathrm{k}}^{2}=1\\ \text{or \hspace{0.17em}}\mathrm{k}=1\\ \text{On putting this value of}\mathrm{k}\text{in equation (2), we get}\\ \text{AC}=\text{BC}\\ \text{i.e.,}\angle \text{A}=\angle \text{B}\end{array}$

Q.7

$\text{If cot}\mathrm{\theta }=\frac{\text{7}}{8},\text{evaluate : (i)}\frac{\left(1+\mathrm{sin\theta }\right)\left(1-\mathrm{sin\theta }\right)}{\left(1+\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\right)},{\text{(ii) cot}}^{2}\mathrm{\theta }$

Ans

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right triangle, right-angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{cot}\mathrm{\theta }=\frac{7}{8}\\ \text{or}\frac{\text{AB}}{\text{BC}}=\frac{7}{8}\\ \text{Let BC be 8k. Therefore, AB is 7k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{or AC}}^{2}={\left(\text{7k}\right)}^{2}+{\left(8\mathrm{k}\right)}^{2}\\ {\text{or AC}}^{2}=49{\mathrm{k}}^{2}+64{\mathrm{k}}^{2}\\ {\text{or \hspace{0.17em}AC}}^{2}=11\text{3}{\mathrm{k}}^{2}\\ \text{or \hspace{0.17em}AC}=\sqrt{113}\mathrm{k}\\ \mathrm{sin\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{8\mathrm{k}}{\sqrt{113}\mathrm{k}}=\frac{8}{\sqrt{113}}\\ \mathrm{cos\theta }=\frac{\text{side adjacent to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{7\mathrm{k}}{\sqrt{113}\mathrm{k}}=\frac{7}{\sqrt{113}}\\ \text{(i)}\frac{\left(1+\mathrm{sin\theta }\right)\left(1-\mathrm{sin\theta }\right)}{\left(1+\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\right)}=\frac{1-{\mathrm{sin}}^{2}\mathrm{\theta }}{1-{\mathrm{cos}}^{2}\mathrm{\theta }}=\frac{1-{\left(\frac{8}{\sqrt{113}}\right)}^{2}}{1-{\left(\frac{7}{\sqrt{113}}\right)}^{2}}=\frac{113-64}{113-49}=\frac{49}{64}\\ {\text{(ii) cot}}^{2}\mathrm{\theta }={\left(\frac{7}{8}\right)}^{2}=\frac{49}{64}\end{array}$

Q.8

$\text{If 3 cot A}=\text{4, check whether}\frac{1-{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{tan}}^{2}\text{A}}={\mathrm{cos}}^{2}\text{A}-{\mathrm{sin}}^{2}\text{A or not.}$

Ans

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right triangle, right-angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{cot A}=\frac{4}{3}\\ \text{or}\frac{\text{AB}}{\text{BC}}=\frac{4}{3}\\ \text{Let BC be 3k. Therefore, AB = 4k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{or AC}}^{2}={\left(\text{4k}\right)}^{2}+{\left(3\mathrm{k}\right)}^{2}\\ {\text{or AC}}^{2}=16{\mathrm{k}}^{2}+9{\mathrm{k}}^{2}\\ {\text{or \hspace{0.17em}AC}}^{2}=25{\mathrm{k}}^{2}\\ \text{or \hspace{0.17em}AC}=5\mathrm{k}\\ \mathrm{sin}\text{A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{3\mathrm{k}}{5\mathrm{k}}=\frac{3}{5}\\ \mathrm{cos}\text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{4\mathrm{k}}{5\mathrm{k}}=\frac{4}{5}\\ \mathrm{tan}\text{A}=\frac{\text{side opposite to angle A}}{\text{side adjacent to angle A}}=\frac{\text{BC}}{\text{AB}}=\frac{3\mathrm{k}}{4\mathrm{k}}=\frac{3}{4}\\ \text{Now,}\\ \frac{1-{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{tan}}^{2}\text{A}}=\frac{1-{\left(\frac{3}{4}\right)}^{2}}{1+{\left(\frac{3}{4}\right)}^{2}}=\frac{7}{25}\\ {\mathrm{cos}}^{2}\text{A}-{\mathrm{sin}}^{2}\text{A}={\left(\frac{4}{5}\right)}^{2}-{\left(\frac{3}{5}\right)}^{2}=\frac{7}{25}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1-{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{tan}}^{2}\text{A}}={\mathrm{cos}}^{2}\text{A}-{\mathrm{sin}}^{2}\text{A}\end{array}$

Q.9

$\begin{array}{l}\text{In triangle ABC, right-angled at B, if tan A}=\frac{1}{\sqrt{3}},\text{find the value of:}\\ \text{(i) sin A cos C}+\text{cos A sin C}\\ \text{(ii) cos A cos C}-\text{sin A sin C}\end{array}$

Ans

$\begin{array}{l}\text{Given that,}\\ \text{tan A}=\frac{1}{\sqrt{3}}\\ \text{or}\frac{\text{BC}}{\text{AB}}=\frac{1}{\sqrt{3}}\\ \text{Let BC be}\mathrm{k}\text{. Then, AB =}\sqrt{3}\mathrm{k}\text{, where}\mathrm{k}\text{is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{or AC}}^{2}={\left(\sqrt{3}\text{k}\right)}^{2}+{\left(\mathrm{k}\right)}^{2}\\ {\text{or AC}}^{2}=3{\mathrm{k}}^{2}+{\mathrm{k}}^{2}\\ {\text{or \hspace{0.17em}AC}}^{2}=4{\mathrm{k}}^{2}\\ \text{or \hspace{0.17em}AC}=2\mathrm{k}\\ \mathrm{sin}\text{A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\mathrm{k}}{2\mathrm{k}}=\frac{1}{2}\\ \mathrm{cos}\text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}\mathrm{k}}{2\mathrm{k}}=\frac{\sqrt{3}}{2}\\ \mathrm{sin}\text{C}=\frac{\text{side opposite to angle C}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}\mathrm{k}}{2\mathrm{k}}=\frac{\sqrt{3}}{2}\\ \mathrm{cos}\text{C}=\frac{\text{side adjacent to angle C}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\mathrm{k}}{2\mathrm{k}}=\frac{1}{2}\\ \text{Now,}\\ \text{(i)}\mathrm{sin}\text{A}\mathrm{cos}\text{C}+\mathrm{cos}\text{A}\mathrm{sin}\text{C}=\frac{1}{2}×\frac{1}{2}+\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}=1\\ \text{(ii)}\mathrm{cos}\text{A}\mathrm{cos}\text{C}-\mathrm{sin}\text{A}\mathrm{sin}\text{C}=\frac{\sqrt{3}}{2}×\frac{1}{2}-\frac{1}{2}×\frac{\sqrt{3}}{2}=0\end{array}$

Q.10

$\begin{array}{l}\text{In }\mathrm{\Delta }\text{\hspace{0.17em}PQR, right-angled at Q, PR}+\text{QR}=\text{25 cm and PQ}=\text{5 cm.}\\ \text{Determine the values of sin\hspace{0.17em}P, cos\hspace{0.17em}P and tan\hspace{0.17em}\hspace{0.17em}P.}\end{array}$

Ans

$\begin{array}{l}\text{Given that,}\\ \text{PR}+\text{QR}=\text{25 cm}\\ \text{and}\\ \text{PQ}=\text{5 cm}\\ \text{Let PR be}\mathrm{x}\text{. Then, QR =}25-\mathrm{x}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{\hspace{0.17em}PQR, we get}\\ {\text{PR}}^{2}={\text{PQ}}^{2}+{\text{QR}}^{2}\\ \text{or}{\mathrm{x}}^{2}={\left(5\right)}^{2}+{\left(25-\mathrm{x}\right)}^{2}\\ \text{or}{\mathrm{x}}^{2}=25+625+{\mathrm{x}}^{2}-50\mathrm{x}\\ \text{or \hspace{0.17em}}50\mathrm{x}=650\\ \text{or \hspace{0.17em}}\mathrm{x}=13\\ \therefore \text{PR}=\mathrm{x}=13\text{cm}\\ \text{and}\\ \text{QR}=25-\mathrm{x}=25-13=12\text{cm}\\ \mathrm{sin}\text{\hspace{0.17em}P}=\frac{\text{side opposite to angle P}}{\text{hypotenuse}}=\frac{\text{QR}}{\text{PR}}=\frac{12}{13}\\ \mathrm{cos}\text{\hspace{0.17em}P}=\frac{\text{side adjacent to angle P}}{\text{hypotenuse}}=\frac{\text{PQ}}{\text{PR}}=\frac{5}{13}\\ \mathrm{tan}\text{\hspace{0.17em}P}=\frac{\text{side opposite to angle P}}{\text{side adjacent to angle P}}=\frac{\text{QR}}{\text{PQ}}=\frac{12}{5}\end{array}$

Q.11

$\begin{array}{l}\text{State whether the following are true or false. Justify}\\ \text{your answer.}\\ \text{(i) The value of tan A is always less than 1.}\\ \text{(ii) sec A}=\frac{12}{5}\text{ for some value of angle A.}\\ \text{(iii) cos A is the abbreviation used for the cosecant}\\ \text{of angle A.}\\ \text{(iv) cot A is the product of cot and A.}\\ \text{(v) sin }\mathrm{\theta }=\frac{4}{3}\text{ for some angle }\mathrm{\theta }.\end{array}$

Ans
(i) Let us consider the following Δ ABC, right-angled at B.

$\begin{array}{l}\text{}\\ \mathrm{tan}\text{\hspace{0.17em}A}=\frac{\text{side opposite to angle A}}{\text{side adjacent to angle A}}=\frac{12}{5}>1\\ \therefore \mathrm{tan}\text{\hspace{0.17em}A}>1\\ \text{So, tan\hspace{0.17em}A<1 is not always true.}\\ \text{Hence, the given statement is false.}\end{array}$

(ii) Let us consider the following Δ ABC, right-angled at B.

$\begin{array}{l}\text{We consider on the above}\mathrm{\Delta }\text{\hspace{0.17em}ABC, right-angled at B.}\\ \\ \mathrm{sec}\text{\hspace{0.17em}A}=\frac{\text{hypotenuse}}{\text{side adjacent to angle A}}=\frac{\text{AC}}{\text{AB}}=\frac{12}{5}\\ \text{Now, by using Pythagoras theorem, we have}\\ {\text{BC}}^{2}={\text{AC}}^{2}-{\text{AB}}^{2}=144-25=119\\ \text{or BC}=\sqrt{119}=10.9\\ \text{Therefore,}\mathrm{sec}\text{\hspace{0.17em}A}=\frac{12}{5}\text{is possible for some values}\\ \text{of angle A.}\\ \text{Hence, the given statement is true.}\\ \text{(iii)}\\ \text{The abbreviation used for the cosecant of angle A is cosec\hspace{0.17em}A.}\\ \text{Therefore, the given statement is false.}\\ \text{(iv)}\\ \text{cot A is not the product of cot and A. Therefore, the given}\\ \text{statement is true.}\\ \text{(v)}\\ \mathrm{sin}\text{\hspace{0.17em}}\mathrm{\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}\\ \text{In a right-angled triangle hypotenuse is the longest side.}\\ \text{Thus,}\\ \mathrm{sin}\text{\hspace{0.17em}}\mathrm{\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}<1\\ \text{Therefore,}\mathrm{sin}\text{\hspace{0.17em}}\mathrm{\theta }=\frac{4}{3}>1\text{is not possible for any value of}\mathrm{\theta }.\\ \text{Hence, the given statement is false.}\end{array}$

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