# NCERT Solutions Class 10 Maths Chapter 8

## NCERT Solutions Class 10 Mathematics Chapter 8: Introduction to Trigonometry

Chapter 8 in the Class 10 CBSE syllabus introduces students to the concept of Trigonometry. Students can subscribe to  NCERT Solutions Class 10 Mathematics Chapter 8 by Extramarks so that they can study from it independently without any assistance from teachers or parents. It will ensure that even the minutest doubts are resolved. In fact, the students would be interested in learning and mastering the topic with ease.. The subject matter experts have written the answers in the solutions while ensuring that they are accurate and adhere to  the guidelines laid down by CBSE.

## NCERT Solutions for Class 10 Mathematics Chapter 8 –

Trigonometry can be a very challenging topic to master for students since it involves so many different formulas and theorems that students need to know and apply them correctly. Furthermore, there are many different kinds of problems that can be asked from this topic. NCERT Solutions for Class 10 Mathematics Chapter 8 will, therefore, be an extremely useful resource as students will find step-by-step explanations to all the problems given in the NCERT textbook.

## NCERT Solutions for Class 10 Mathematics

Students can also access detailed solutions to all the chapters covered in the Class 10 Mathematics textbook listed below:

Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3: Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progressions

Chapter 6: Triangles

Chapter 7: Coordinate Geometry

Chapter 8: Introduction to Trigonometry

Chapter 9: Some Applications of Trigonometry

Chapter 10: Circles

Chapter 11: Constructions

Chapter 12: Areas Related to Circles

Chapter 13: Surface Areas and Volumes

Chapter 14: Statistics

Chapter 15: Probability

## About the Chapter – Introduction to Trigonometry

Trigonometry is a Mathematics concept that focuses on triangles. Trigonometry is a blend of ‘trigono’ and ‘metry’, which means triangle and measure.

The chapter includes proper methods and formulae to help the students find the missing sides and angle of a triangle.  Class 10 chapter 8 has been segmented into five sections.

The first section presents the problems depicted using the right-angle triangle and the second section comprises an introduction to trigonometric ratios with examples. The students can work on the exercises to understand the topics in detail. The segment also explains the derivation of sine, cosine, and several other trigonometric functions.

The third section talks about the trigonometric ratios of measurement. The fourth section presents a few solved examples, trigonometric ratio criteria for the complementary angles and the exercises. The last section explains the subject associated with the trigonometric identities, accompanied by examples and exercises.

## Benefits of NCERT Solutions for Class 10 Mathematics

NCERT Solutions for Class 10 Mathematics can be extremely beneficial for the students to master the topic and increase their confidence in achieving a higher grade.

The multiple advantages of studying NCERT solutions are:

• The solutions have been created by experienced faculty and subject-matter experts with years of experience thus students can be assured that they are referring to reliable and authentic learning material.
• The NCERT Solutions strictly follow the CBSE Guidelines.
• Important Terms to Remember in Height and Distance
• The solutions break down the complex trigonometric problems in a step by step manner to ease the students’ learning process.
• The solutions are descriptive and to the point. .
• The answers have been described, accompanied by Illustrations and examples.
• The language used in the solutions is easy-to-understand.

## Related Questions

Question: State whether the following condition is true or false: The value of sinθ increases as θ increases between 0 and 90 degrees.

Solution: We know the values of sine for different values of θ between 0 and 90 degrees

sin 0 = 0

sin 30 = 1/2 = 0.5

sin 45 = 1/√2 = 0.7071

sin 60 = √3/2 = 0.866

sin 90 = 1

It can be seen that the value of sinθ increases as θ increases between 0 and 90 degrees. Therefore, the given statement is true.

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

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

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

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

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.

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

Q.12

$\begin{array}{l}\text{Evaluate the following:}\\ \text{(i) sin\hspace{0.17em}60°\hspace{0.17em}cos 30°}+\text{sin 30°\hspace{0.17em}cos 60°}\\ {\text{(ii) 2tan}}^{\text{2}}\text{45°}+{\text{cos}}^{\text{2}}\text{30°}-{\text{sin}}^{\text{2}}\text{60°}\\ \text{(iii)}\frac{\text{cos 45°}}{\text{sec 30°}+\text{cosec 30°}}\\ \text{(iv)}\frac{\text{sin 30°}+\text{tan 45°}-\text{cosec 60°}}{\text{sec 30°}+\text{cos 60°}+\text{cot 45°}}\\ \text{(v)\hspace{0.17em}\hspace{0.17em}}\frac{{\text{5cos}}^{2}\text{60°}+{\text{4sec}}^{2}\text{30°}-{\text{tan}}^{2}\text{45°}}{{\text{sin}}^{2}\text{30°}+{\text{cos}}^{2}\text{30°}}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{sin\hspace{0.17em}60°\hspace{0.17em}cos 30°}+\text{sin 30°\hspace{0.17em}cos 60°}=\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}+\frac{1}{2}×\frac{1}{2}=1\\ \text{(ii)}\\ {\text{2tan}}^{\text{2}}\text{45°}+{\text{cos}}^{\text{2}}\text{30°}-{\text{sin}}^{\text{2}}\text{60°}=2×{\left(1\right)}^{2}+{\left(\frac{\sqrt{3}}{2}\right)}^{2}-{\left(\frac{\sqrt{3}}{2}\right)}^{2}=2\\ \text{(iii)}\\ \frac{\text{cos 45°}}{\text{sec 30°}+\text{cosec 30°}}=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\sqrt{3}}{2\sqrt{2}\left(1+\sqrt{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}\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}\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}}=\frac{\sqrt{3}}{2\sqrt{2}\left(1+\sqrt{3\right)}}×\frac{\left(1-\sqrt{3\right)}}{\left(1-\sqrt{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}\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}\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}}=\frac{\sqrt{3}-3}{2\sqrt{2}\left(1-3\right)}=\frac{3-\sqrt{3}}{4\sqrt{2}}\\ \text{(iv)}\\ \frac{\text{sin 30°}+\text{tan 45°}-\text{cosec 60°}}{\text{sec 30°}+\text{cos 60°}+\text{cot 45°}}=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{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}\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}\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}\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}}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{3}{2}}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}\\ \text{}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}×\frac{3\sqrt{3}-4}{3\sqrt{3}-4}\\ \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}\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}\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(3\sqrt{3}-4\right)}^{2}}{{\left(3\sqrt{3}\right)}^{2}-16}=\frac{{\left(3\sqrt{3}\right)}^{2}-24\sqrt{3}+16}{27-16}\\ \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}\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}\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{43-24\sqrt{3}}{11}\\ \text{(v)\hspace{0.17em}\hspace{0.17em}}\\ \frac{{\text{5cos}}^{2}\text{60°}+{\text{4sec}}^{2}\text{30°}-{\text{tan}}^{2}\text{45°}}{{\text{sin}}^{2}\text{30°}+{\text{cos}}^{2}\text{30°}}=\frac{5×{\left(\frac{1}{2}\right)}^{2}+4×{\left(\frac{2}{\sqrt{3}}\right)}^{2}-1}{{\left(\frac{1}{2}\right)}^{2}+{\left(\frac{\sqrt{3}}{2}\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}\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}\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}\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}\hspace{0.17em}\hspace{0.17em}}=\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{15+64-12}{12}}{\frac{4}{4}}=\frac{67}{12}\end{array}$

Q.13

$\begin{array}{l}\text{Choose the correct option and justify your choice:}\\ \text{(i)}\frac{\text{2tan30°}}{\text{1}+{\text{tan}}^{2}\text{30°}}=\\ \text{(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°}\\ \text{(ii)}\frac{\text{1}-{\text{tan}}^{2}\text{45°}}{\text{1}+{\text{tan}}^{2}\text{45°}}=\\ \text{(A) tan 90° (B) 1 (C) sin 45° (D) 0}\\ \text{(iii) sin2A}=\text{2sinA is true when A}=\\ \text{(A) 0° (B) 30° (C) 45° (D) 60°}\\ \text{(iv)}\frac{\text{2tan30°}}{1-{\text{tan}}^{2}\text{30°}}=\\ \text{(A) cos 60° (B)sin 60° (C) tan 60° (D) sin 30}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \frac{\text{2tan30°}}{\text{1}+{\text{tan}}^{2}\text{30°}}=\frac{2×\frac{1}{\sqrt{3}}}{1+{\left(\frac{1}{\sqrt{3}}\right)}^{2}}=\frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}}=\frac{2×3}{4\sqrt{3}}=\frac{\sqrt{3}}{2}=\mathrm{sin}60\mathrm{°}\\ \text{Hence, the correct option is (A).}\\ \text{(ii)}\\ \text{}\frac{\text{1}-{\text{tan}}^{2}\text{45°}}{\text{1}+{\text{tan}}^{2}\text{45°}}=\frac{1-1}{1+1}=0\\ \text{Hence, the correct option is (D).}\\ \text{(iii)}\\ \text{sin2A}=\text{2sinA}\\ \text{On putting A}=0\mathrm{°},\text{we get}\\ \text{sin\hspace{0.17em}}0\mathrm{°}=\text{2sin\hspace{0.17em}}0\mathrm{°}\\ \text{or 0}=2×0\\ \text{or 0}=0\\ \text{Therefore, the correct option is (A).}\\ \text{Other given options does not satisfy the given condition.}\\ \text{(iv)}\\ \frac{\text{2tan30°}}{1-{\text{tan}}^{2}\text{30°}}=\frac{\text{2}×\frac{1}{\sqrt{3}}}{1-{\left(\frac{1}{\sqrt{3}}\right)}^{2}}=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}=\frac{2}{\sqrt{3}}×\frac{3}{2}=\sqrt{3}=\mathrm{tan}\text{\hspace{0.17em}}60\mathrm{°}\\ \text{Therefore, the correct option is (C).}\end{array}$

Q.14

$\text{If tan}\left(\mathrm{A}+\mathrm{B}\right)=\sqrt{\text{3}}\text{â€„and tan}\left(\mathrm{A}-\mathrm{B}\right)=\frac{1}{\sqrt{\text{3}}}\text{; 0° < A + B}\le \text{90°; A > B, find A and B.}$

Ans.

$\begin{array}{l}\text{We have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} tan}\left(\mathrm{A}+\mathrm{B}\right)=\sqrt{\text{3}}\\ \text{or tan (A}+\text{B}\right)=\text{tan\hspace{0.17em}60°}\\ \text{or A}+\text{B}=60\mathrm{°}\text{}...\text{(1)}\\ \text{Again, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan}\left(\mathrm{A}-\mathrm{B}\right)=\frac{1}{\sqrt{\text{3}}}\\ \text{or tan (A}-\text{B}\right)=\text{tan\hspace{0.17em}}3\text{0°}\\ \text{or A}-\text{B}=30\mathrm{°}\text{}...\text{(2)}\\ \text{On adding equations (1) and (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 2A}=90\mathrm{°}\\ \text{or A}=45\mathrm{°}\\ \text{On putting this value of A in equation (1), we get}\\ \text{B}=15\mathrm{°}\end{array}$

Q.15

$\begin{array}{l}\text{State whether the following are true or false.}\\ \text{Justify your answer.}\\ \text{(i) sin}\left(\mathrm{A}+\mathrm{B}\right)=\text{sin A}+\text{sin B.}\\ \text{(ii) The value of sin}\mathrm{\theta }\text{â€„increases asâ€„}\mathrm{\theta }\text{â€„increases.}\\ \text{(iii) The value of cos}\mathrm{\theta }\text{â€„increases asâ€„}\mathrm{\theta }\text{â€„increases.}\\ \text{(iv) sin}\mathrm{\theta }=\text{cos}\mathrm{\theta }\text{â€„for all values ofâ€„}\mathrm{\theta }.\\ \text{(v) cot A is not defined for A}=0.\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Let A}=30\mathrm{°}\text{and B}=60\mathrm{°}.\\ \text{Now,}\\ \mathrm{sin}\left(\text{A}+\text{B)}=\mathrm{sin}\left(30\mathrm{°}+60\mathrm{°}\text{)}=\mathrm{sin}90\mathrm{°}=1\\ \text{and}\\ \text{sin A}+\text{sin B}=\text{sin}30\mathrm{°}+\text{sin\hspace{0.17em}\hspace{0.17em}}60\mathrm{°}=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}\\ \therefore \mathrm{sin}\left(\text{A}+\text{B)}\ne \text{sin A}+\text{sin B}\\ \text{Hence, the given statement is false.}\\ \text{(ii)}\\ \text{We know that}\\ \mathrm{sin}0\mathrm{°}=0,\\ \mathrm{sin}30\mathrm{°}=\frac{1}{2}=0.5,\\ \mathrm{sin}45\mathrm{°}=\frac{1}{\sqrt{2}}=0.7,\\ \mathrm{sin}60\mathrm{°}=\frac{\sqrt{3}}{2}=0.87,\\ \mathrm{sin}90\mathrm{°}=1\\ \text{We see that the value of sin}\mathrm{\theta }\text{increases as}\mathrm{\theta }\text{increases}\\ \text{in the interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{Hence, the given statement is true if}\mathrm{\theta }\text{lies in the}\\ \text{interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{(iii)}\\ \text{We know that}\\ \mathrm{cos}0\mathrm{°}=1,\\ \mathrm{cos}30\mathrm{°}=\frac{\sqrt{3}}{2}=0.87,\\ \mathrm{cos}45\mathrm{°}=\frac{1}{\sqrt{2}}=0.7,\\ \mathrm{cos}60\mathrm{°}=\frac{1}{2}=0.5,\\ \mathrm{cos}90\mathrm{°}=0\\ \text{We see that the value of}\mathrm{cos\theta }\text{decreases as}\mathrm{\theta }\text{increases}\\ \text{in the interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{Hence, the given statement is false if}\mathrm{\theta }\text{lies in the}\\ \text{interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{(iv)}\\ \text{We know that}\\ \mathrm{sin}0\mathrm{°}=0\text{and}\mathrm{cos}0\mathrm{°}=1,\\ \mathrm{sin}30\mathrm{°}=\frac{1}{2}=0.5\text{and}\mathrm{cos}30\mathrm{°}=\frac{\sqrt{3}}{2}=0.87.\\ \text{Therefore, sin}\mathrm{\theta }\ne \text{cos}\mathrm{\theta }\text{for all values of}\mathrm{\theta }.\\ \text{Hence, the given statement is false}.\\ \text{(v)}\\ \text{We know that cot A}=\frac{\mathrm{cos}\text{A}}{\mathrm{sin}\text{A}}.\text{}\\ \text{So, for A}=0,\text{we have}\\ \text{cot 0}=\frac{\mathrm{cos}\text{0}}{\mathrm{sin}\text{0}}=\frac{1}{0}\\ \text{We know that division by 0 is not defined. Therefore, cot 0}\\ \text{is not defined.}\\ \text{Hence, the given statement is true}.\end{array}$

Q.16

$\begin{array}{l}\text{Evaluate\hspace{0.17em}:}\\ \text{(i)}\frac{\text{sin 18°}}{\text{cos 72}°}\text{}\\ \text{(ii)}\frac{\text{tan 26°}}{\text{cot 64°}}\text{}\\ \text{(iii) cos 48°}-\text{sin 42°}\\ \text{(iv) cosec 31°}-\text{sec 59°}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{}\frac{\text{sin 18°}}{\text{cos 72}\mathrm{°}}=\frac{\text{cos(90°}-18\text{°}\right)}{\text{cos 72}\mathrm{°}}=\frac{\text{cos 72}\mathrm{°}}{\text{cos 72}\mathrm{°}}=1\\ \\ \text{(ii)}\\ \text{}\frac{\text{tan 26°}}{\text{cot 64°}}=\frac{\text{cot (90°}-\text{26°)}}{\text{cot 64°}}=\frac{\text{cot 64°}}{\text{cot 64°}}=1\\ \\ \text{(iii)}\\ \text{cos 48°}-\text{sin 42°}=\text{sin (90°}-\text{48°)}-\text{sin 42°}\\ \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}\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{sin 42°}-\text{sin 42°}\\ \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}\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0\\ \text{(iv)}\\ \text{cosec 31°}-\text{sec 59°}=\text{sec (90°}-\text{31°)}-\text{sec 59°}\\ \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}\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}\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}}=\text{sec 59°}-\text{sec 59°}\\ \text{\hspace{0.17em}\hspace{0.17em}}=0\end{array}$

Q.17

$\begin{array}{l}\text{Show that\hspace{0.17em}:}\\ \text{(i) tan 48° tan 23° tan 42° tan 67°}=1\\ \text{(ii) cos 38° cos 52°}-\text{sin 38° sin 52°}=0\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{tan 48° tan 23° tan 42° tan 67°}\\ =\text{cot (90°}-\text{48°) cot (90°}-23\text{°) tan 42° tan 67°}\\ =\text{cot}42°\text{\hspace{0.17em} cot 67°\hspace{0.17em}\hspace{0.17em}tan 42°\hspace{0.17em}\hspace{0.17em}tan 67°}\\ =\text{cot}42°\text{\hspace{0.17em}tan 42°\hspace{0.17em}\hspace{0.17em}tan 67° cot 67°}\\ =1×1=1\\ \text{(ii)}\\ \text{cos 38° cos 52°}-\text{sin 38° sin 52°}\\ =\mathrm{sin}\text{(90°}-3\text{8°)\hspace{0.17em}cos (90°}-3\text{8°)}-\text{sin 38° sin 52°}\\ =\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}52°\hspace{0.17em}}\mathrm{sin}38\text{°}-\text{sin 38° sin 52°}\\ =0\end{array}$

Q.18

$\text{If tan 2A}=\text{cot}\left(\text{A}-18\mathrm{°}\right),\text{where 2A is an acute angle},\text{find the value of A}.$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{tan 2A}=\text{cot}\left(\text{A}-18\mathrm{°}\right)\\ \text{or cot (90°}-2\text{A}\right)=\text{cot}\left(\text{A}-18\mathrm{°}\right)\\ \text{or 90°}-2\text{A}=\text{A}-18\mathrm{°}\\ \text{or}3\text{A}=108\mathrm{°}\\ \text{or A}=\frac{108\mathrm{°}}{3}=36\mathrm{°}\end{array}$

Q.19

$\text{If tan A}=\text{cot B, prove that A}+\text{B}=\text{90°.}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} tan A}=\text{cot B}\\ \text{or cot (90°}-\text{A)}=\mathrm{cot}\text{\hspace{0.17em}\hspace{0.17em}B}\\ \text{or 90°}-\text{A}=\text{B}\\ \text{or A}+\text{B}=\text{90°}\end{array}$

Q.20

$\text{If sec 4A}=\text{cosec (A}-\text{20°), where 4A is an acute angle, find the value of A.}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{sec 4A}=\text{cosec (A}-\text{20°)}\\ \text{or cosec (90°}-\text{4A)}=\text{cosec (A}-\text{20°)}\\ \text{or 90°}-\text{4A}=\text{A}-\text{20°}\\ \text{or 5A}=110\text{°}\\ \text{or A}=\frac{110\text{°}}{5}=22\text{°}\end{array}$

Q.21

$\begin{array}{l}\text{If A, B and C are interior angles of a triangle ABC,}\\ \text{then show that}\\ \text{sin}\left(\frac{\text{B+C}}{2}\right)=\mathrm{cos}\frac{\text{A}}{2}.\end{array}$

Ans.

$\begin{array}{l}\text{Given that A, B and C are interior angles of a triangle ABC.}\\ \therefore \text{A}+\text{B}+\text{C}=180\mathrm{°}\\ \text{or A}=180\mathrm{°}-\text{B}-\text{C}\\ \text{Now,}\\ \text{sin}\left(\frac{\text{B}+\text{C}}{2}\right)=\mathrm{cos}\left(90\mathrm{°}-\frac{\text{B}+\text{C}}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cos}\left(\frac{180\mathrm{°}-\text{B}-\text{C}}{2}\right)\\ \text{}=\mathrm{cos}\left(\frac{\text{A}}{2}\right)\text{}\end{array}$

Q.22

$\text{Express sin 67°}+\text{cos 75° in terms of trigonometric ratios of angles between 0° and 45°.}$

Ans.

$\begin{array}{l}\text{sin 67°}+\text{cos 75°}=\mathrm{cos}\text{(90°}-\text{67°)}+\text{sin (90°}-\text{75°)}\\ \text{}=\mathrm{cos}23\text{°}+\text{sin 15°}\end{array}$

Q.23

$\text{Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.}$

Ans.

$\begin{array}{l}\text{We know that}\\ {\text{cosec}}^{\text{2}}\text{\hspace{0.17em}A}=1+{\text{cot}}^{2}\text{\hspace{0.17em}A}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{sin}}^{2}\text{A}}=1+{\text{cot}}^{2}\text{\hspace{0.17em}A}\\ \text{or \hspace{0.17em}\hspace{0.17em}}{\mathrm{sin}}^{2}\text{A}=\frac{1}{1+{\text{cot}}^{2}\text{\hspace{0.17em}A}}\\ \text{or \hspace{0.17em}}\mathrm{sin}\text{A}=\frac{1}{\sqrt{1+{\text{cot}}^{2}\text{\hspace{0.17em}A}}}\\ \text{Also, we know that}\\ {\text{sec}}^{\text{2}}\text{\hspace{0.17em}A}=1+{\text{tan}}^{2}\text{\hspace{0.17em}A}\\ {\text{or sec}}^{\text{2}}\text{\hspace{0.17em}A}=1+\frac{1}{{\text{cot}}^{2}\text{\hspace{0.17em}A}}\\ \text{or}\mathrm{sec}\text{\hspace{0.17em}A}=\sqrt{\frac{1+{\mathrm{cot}}^{2}\text{A}}{{\mathrm{cot}}^{2}\text{A}}}=\frac{\sqrt{1+{\mathrm{cot}}^{2}\text{A}}}{\mathrm{cot}\text{A}}\\ \text{Also, we know that}\\ \text{tan\hspace{0.17em}A}=\frac{\mathrm{sin}\text{A}}{\mathrm{cos}\text{A}}=\frac{1}{\frac{\mathrm{cos}\text{A}}{\mathrm{sin}\text{A}}}=\frac{1}{\mathrm{cot}\text{A}}\end{array}$

Q.24

$\text{Write all the other trigonometric ratios of}\angle \text{A in terms of sec A.}$

Ans.

$\begin{array}{l}\text{We know that}\\ {\text{sin}}^{\text{2}}\text{\hspace{0.17em}A}=1-{\text{cos}}^{2}\text{\hspace{0.17em}A}\\ {\text{or \hspace{0.17em}\hspace{0.17em}sin}}^{\text{2}}\text{\hspace{0.17em}A}=1-\frac{1}{{\text{sec}}^{2}\text{\hspace{0.17em}A}}\\ \text{or \hspace{0.17em}\hspace{0.17em}}{\mathrm{sin}}^{2}\text{A}=\frac{{\text{sec}}^{2}\text{\hspace{0.17em}A}-1}{{\text{sec}}^{2}\text{\hspace{0.17em}A}}\\ \text{or \hspace{0.17em}}\mathrm{sin}\text{A}=\frac{\sqrt{{\text{sec}}^{2}\text{\hspace{0.17em}A}-1}}{\text{sec\hspace{0.17em}A}}\\ \mathrm{or}\text{}\frac{1}{\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{A}}=\frac{\sqrt{{\text{sec}}^{2}\text{\hspace{0.17em}A}-1}}{\text{sec\hspace{0.17em}A}}\\ \mathrm{or}\text{}\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{A}=\frac{\text{sec\hspace{0.17em}A}}{\sqrt{{\text{sec}}^{2}\text{\hspace{0.17em}A}-1}}\\ \text{Also, we know that}\\ \text{sec\hspace{0.17em}A cos\hspace{0.17em}A}=1\\ \text{or cos\hspace{0.17em}A}=\frac{1}{\text{sec\hspace{0.17em}A}}\\ \\ \text{Also, we know that}\\ {\text{sec}}^{2}\text{A}-{\text{tan}}^{2}\text{\hspace{0.17em}A}=1\\ {\text{or tan}}^{2}\text{\hspace{0.17em}A}={\text{sec}}^{2}\text{A}-1\\ \mathrm{or}\text{tan\hspace{0.17em}A}=\sqrt{{\text{sec}}^{2}\text{A}-1}\\ \text{Also, we know that}\\ \text{tan\hspace{0.17em}A cot\hspace{0.17em}A}=1\\ \mathrm{or}\text{cot\hspace{0.17em}A}=\frac{1}{\text{tan\hspace{0.17em}A}}=\frac{1}{\sqrt{{\text{sec}}^{2}\text{A}-1}}\end{array}$

Q.25

$\begin{array}{l}\text{Evaluate:}\\ \text{(i)}\frac{{\text{sin}}^{2}\text{63°}+{\text{sin}}^{2}\text{27°}}{{\text{cos}}^{2}\text{17°}+{\text{cos}}^{2}\text{73°}}\\ \text{(ii) sin 25° cos 65°}+\text{cos 25° sin 65°}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{}\frac{{\text{sin}}^{2}\text{63°}+{\text{sin}}^{2}\text{27°}}{{\text{cos}}^{2}\text{17°}+{\text{cos}}^{2}\text{73°}}=\frac{{\text{sin}}^{2}\text{\hspace{0.17em}63°}+\mathrm{co}{\text{s}}^{2}\text{(90°}-\text{27°)}}{{\text{sin}}^{2}\text{(90°}-\text{17°)}+{\text{cos}}^{2}\text{\hspace{0.17em}73°}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\text{sin}}^{2}\text{\hspace{0.17em}63°}+\mathrm{co}{\text{s}}^{2}\text{\hspace{0.17em}63°}}{{\text{sin}}^{2}\text{\hspace{0.17em}73°}+{\text{cos}}^{2}\text{\hspace{0.17em}73°}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{1}=1\\ \text{(ii)}\\ \text{sin\hspace{0.17em}25° cos 65°}+\text{cos\hspace{0.17em}25° sin\hspace{0.17em}65°}\\ =\text{sin\hspace{0.17em}25°\hspace{0.17em}sin(90°}-\text{65°}\right)+\text{cos 25° cos(90°}-\text{\hspace{0.17em}65°)}\\ =\text{sin\hspace{0.17em}25° sin\hspace{0.17em}25°}+\text{cos 25°\hspace{0.17em}cos 25°}\\ ={\text{sin}}^{2}\text{\hspace{0.17em}25°}+{\text{cos}}^{2}\text{25°}\\ =1\end{array}$

Q.26

$\begin{array}{l}\text{Choose the correct option. Justify your choice.}\\ {\text{(i) 9 sec}}^{2}\text{A}-{\text{9 tan}}^{2}\text{A}=\\ \text{(A) 1 (B) 9 (C) 8 (D) 0}\\ \text{(ii) (1 + tan}\mathrm{\theta }+\text{sec}\mathrm{\theta }\text{) (1 + cot}\mathrm{\theta }-\text{cosec}\mathrm{\theta }\text{)}=\\ \text{(A) 0 (B) 1 (C) 2 (D)}-\text{1}\\ \text{(iii) (sec A + tan A) (1}-\text{sin A)}=\\ \text{(A) sec A (B) sin A (C) cosec A (D) cos A}\\ \text{(iv)}\frac{1+{\mathrm{tan}}^{2}\mathrm{A}}{1+{\mathrm{cot}}^{2}\text{}\mathrm{A}}=\\ {\text{(A) sec}}^{2}\mathrm{A}\text{(B)}-{\text{1 (C) cot}}^{2}{\text{A (D) tan}}^{2}\mathrm{A}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ {\text{9 sec}}^{2}\text{A}-{\text{9 tan}}^{2}\text{A}=9\left({\text{sec}}^{2}\text{A}-{\text{tan}}^{2}\text{A)}=9×1=9\\ \text{Therefore, the correct option is (B).}\\ \text{(ii)}\\ \text{(1}+\text{tan}\mathrm{\theta }+\text{sec}\mathrm{\theta }\text{) (}1+\mathrm{cot\theta }-\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{\theta }\text{)}\\ =\left(1+\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}+\frac{1}{\mathrm{cos\theta }}\right)\left(1+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}-\frac{1}{\mathrm{sin\theta }}\right)\\ =\left(\frac{\mathrm{sin\theta }+\mathrm{cos\theta }+1}{\mathrm{cos\theta }}\right)\left(\frac{\mathrm{sin\theta }+\mathrm{cos\theta }-1}{\mathrm{sin\theta }}\right)\\ =\frac{{\left(\mathrm{sin\theta }+\mathrm{cos\theta }\right)}^{2}-1}{\mathrm{sin\theta cos\theta }}=\frac{{\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }+2\mathrm{sin\theta cos\theta }-1}{\mathrm{sin\theta cos\theta }}\\ =\frac{1+2\mathrm{sin\theta cos\theta }-1}{\mathrm{sin\theta cos\theta }}=2\\ \text{Therefore, the correct option is (C).}\\ \text{(iii)}\\ \text{(sec A + tan A) (1}-\text{sin A)}\\ =\left(\frac{1}{\text{cos A}}+\frac{\mathrm{sin}\text{A}}{\text{cos A}}\right)\left(\text{1}-\text{sin A}\right)\\ =\left(\frac{1+\mathrm{sin}\text{A}}{\text{cos A}}\right)\left(\text{1}-\text{sin A}\right)\\ =\frac{1-{\mathrm{sin}}^{2}\text{A}}{\text{cos A}}=\frac{{\mathrm{cos}}^{2}\text{A}}{\text{cos A}}=\mathrm{cos}\text{A}\\ \text{Therefore, the correct option is (D).}\\ \text{(iv)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1+{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{cot}}^{2}\text{A}}=\frac{{\mathrm{sec}}^{2}\text{A}}{{\mathrm{cosec}}^{2}\text{A}}=\frac{{\mathrm{sin}}^{2}\text{A}}{{\mathrm{cos}}^{2}\text{A}}={\mathrm{tan}}^{2}\text{A}\\ \text{Therefore, the correct option is (D).}\end{array}$

Q.27

$\begin{array}{l}\text{Prove the following identities, where the angles involved}\\ \text{are acute angles for which the expressions are defined.}\\ \text{(i) (cosec}\mathrm{\theta }-\text{cot}\mathrm{\theta }{\right)}^{2}=\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}\\ \text{(ii)}\frac{\mathrm{cos}\text{A}}{1+\mathrm{sin}\text{A}}+\frac{1+\mathrm{sin}\text{A}}{\mathrm{cos}\text{A}}=2\mathrm{sec}\text{A}\\ \text{(iii)}\frac{\mathrm{tan\theta }}{1-\mathrm{cot\theta }}+\frac{\mathrm{cot\theta }}{1-\mathrm{tan\theta }}=1+\mathrm{sec\theta cosec\theta }\\ \text{[Hint :Write the expression in terms of sin}\mathrm{\theta }\text{and cos}\mathrm{\theta }.\right]\\ \text{(iv)}\frac{1+\mathrm{sec}\text{A}}{\mathrm{sec}\text{A}}=\frac{{\mathrm{sin}}^{2}\text{A}}{1-\mathrm{cos}\text{A}}\\ \text{[Hint: Simplify LHS and RHS separately]}\\ \text{(v)}\frac{\mathrm{cos}\text{A}-\mathrm{sin}\text{A}+1}{\mathrm{cos}\text{A}+\mathrm{sin}\text{A}-1}=\text{cosec A}+\text{cot A, using the identity}\\ {\text{cosec}}^{\text{2}}\text{A}=\text{1}+{\text{cot}}^{2}\text{\hspace{0.17em}A.}\\ \text{(vi)}\sqrt{\frac{1+\mathrm{sin}\text{A}}{1-\mathrm{sin}\text{A}}}=\mathrm{sec}\text{A}+\mathrm{tan}\text{A}\\ \text{(vii)}\frac{\text{sin}\mathrm{\theta }-{\text{2sin}}^{3}\mathrm{\theta }}{{\text{2cos}}^{3}\mathrm{\theta }-\text{cos}\mathrm{\theta }}=\text{tan}\mathrm{\theta }\\ {\text{(viii) (sin A + cosec A)}}^{2}+{\left(\text{cos A}+\text{sec A}\right)}^{\text{2}}=7+{\mathrm{tan}}^{2}\text{A}+{\mathrm{cot}}^{2}\text{A}\\ \text{(ix) (cosec A}-\text{sin A) (sec A}-\text{cos A)}=\frac{1}{\mathrm{tan}\text{A}+\mathrm{cot}\text{A}}\\ \text{[Hint: Simplify LHS and RHS separately]}\\ \text{(x)}\frac{1+{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{cot}}^{2}\text{A}}={\left(\frac{\text{1}-\text{tanA}}{\text{1}-\text{cot A}}\right)}^{2}={\mathrm{tan}}^{2}\text{A}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{LHS}={\left(\text{cosec}\mathrm{\theta }-\text{cot}\mathrm{\theta }\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}\hspace{0.17em}}={\mathrm{cosec}}^{2}\mathrm{\theta }+{\mathrm{cot}}^{2}\mathrm{\theta }-2\text{cosec}\mathrm{\theta }\text{\hspace{0.17em}cot}\mathrm{\theta }\\ \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}}=\frac{1}{{\mathrm{sin}}^{2}\mathrm{\theta }}+\frac{{\mathrm{cos}}^{2}\mathrm{\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }}-\frac{2\mathrm{cos\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }}\\ \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}}=\frac{1+{\mathrm{cos}}^{2}\mathrm{\theta }-2\mathrm{cos\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }}\\ \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}}=\frac{\left(1-\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\right)}{1-{\mathrm{cos}}^{2}\mathrm{\theta }}=\frac{\left(1-\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\right)}{\left(1+\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\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-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}=\text{RHS}\\ \\ \text{(ii)}\\ \text{LHS}=\frac{\mathrm{cos}\text{A}}{1+\mathrm{sin}\text{A}}+\frac{1+\mathrm{sin}\text{A}}{\mathrm{cos}\text{A}}\\ \text{}=\frac{{\mathrm{cos}}^{2}\text{A}+{\left(1+\mathrm{sin}\text{A)}}^{2}}{\left(1+\mathrm{sin}\text{A)}\mathrm{cos}\text{A}}\\ \text{}=\frac{{\mathrm{cos}}^{2}\text{A}+{\mathrm{sin}}^{2}\text{A}+2\mathrm{sin}\text{A}+1}{\left(1+\mathrm{sin}\text{A)}\mathrm{cos}\text{A}}\\ \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+1+2\mathrm{sin}\text{A}}{\left(1+\mathrm{sin}\text{A)}\mathrm{cos}\text{A}}\\ \text{}=\frac{2\left(1+\mathrm{sin}\text{A)}}{\left(1+\mathrm{sin}\text{A)}\mathrm{cos}\text{A}}=2\mathrm{sec}\text{A}=\text{RHS}\\ \text{}\\ \text{(iii)}\\ \text{LHS}=\frac{\mathrm{tan\theta }}{1-\mathrm{cot\theta }}+\frac{\mathrm{cot\theta }}{1-\mathrm{tan\theta }}\\ \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}}\\ \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}}=\frac{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}{1-\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}+\frac{\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}{1-\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}\\ \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}}=\frac{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}{\frac{\mathrm{sin\theta }-\mathrm{cos\theta }}{\mathrm{sin\theta }}}+\frac{\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}{\frac{\mathrm{cos\theta }-\mathrm{sin\theta }}{\mathrm{cos\theta }}}\\ \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}}=\frac{{\mathrm{sin}}^{2}\mathrm{\theta }}{\mathrm{cos\theta }\left(\mathrm{sin\theta }-\mathrm{cos\theta }\right)}+\frac{{\mathrm{cos}}^{2}\mathrm{\theta }}{\mathrm{sin\theta }\left(\mathrm{cos\theta }-\mathrm{sin\theta }\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{{\mathrm{sin}}^{3}\mathrm{\theta }-{\mathrm{cos}}^{3}\mathrm{\theta }}{\mathrm{sin\theta cos\theta }\left(\mathrm{sin\theta }-\mathrm{cos\theta }\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left({\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }+\mathrm{sin\theta cos\theta }\right)\left(\mathrm{sin\theta }-\mathrm{cos\theta }\right)}{\mathrm{sin\theta cos\theta }\left(\mathrm{sin\theta }-\mathrm{cos\theta }\right)}\\ \text{}=\frac{1+\mathrm{sin\theta cos\theta }}{\mathrm{sin\theta cos\theta }}=1+\mathrm{sec\theta cosec}\text{\hspace{0.17em}}\mathrm{\theta }=\text{RHS}\\ \text{(iv)}\\ \text{LHS}=\frac{1+\mathrm{sec}\text{A}}{\mathrm{sec}\text{A}}=\frac{1+\frac{1}{\mathrm{cos}\text{A}}}{\frac{1}{\mathrm{cos}\text{A}}}=\frac{1+\mathrm{cos}\text{A}}{1}×\frac{1-\mathrm{cos}\text{A}}{1-\mathrm{cos}\text{A}}\\ \text{}=\frac{1-{\mathrm{cos}}^{2}\text{A}}{1-\mathrm{cos}\text{A}}=\frac{{\mathrm{sin}}^{2}\text{A}}{1-\mathrm{cos}\text{A}}=\text{RHS}\\ \text{(v)}\\ \text{LHS}=\frac{\mathrm{cos}\text{A}-\mathrm{sin}\text{A}+1}{\mathrm{cos}\text{A}+\mathrm{sin}\text{A}-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}\hspace{0.17em}}=\frac{\frac{\mathrm{cos}\text{A}}{\mathrm{sin}\text{A}}-\frac{\mathrm{sin}\text{A}}{\mathrm{sin}\text{A}}+\frac{1}{\mathrm{sin}\text{A}}}{\frac{\mathrm{cos}\text{A}}{\mathrm{sin}\text{A}}+\frac{\mathrm{sin}\text{A}}{\mathrm{sin}\text{A}}-\frac{1}{\mathrm{sin}\text{A}}}\\ \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}}=\frac{\mathrm{cot}\text{A}-1+\mathrm{cosec}\text{A}}{\mathrm{cot}\text{A}+1-\mathrm{cosec}\text{\hspace{0.17em}A}}\\ \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}}=\frac{\mathrm{cot}\text{A}-\left(1-\mathrm{cosec}\text{A)}}{\mathrm{cot}\text{A}+\left(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}×\frac{\mathrm{cot}\text{A}-\left(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}{\mathrm{cot}\text{A}-\left(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}\\ \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}}=\frac{{\mathrm{cot}}^{2}\text{A}+{\left(1-\mathrm{cosec}\text{A)}}^{2}-2\mathrm{cotA}\left(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}{{\mathrm{cot}}^{2}\text{A}-{\left(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}^{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}}=\frac{{\mathrm{cot}}^{2}\text{A}+1+{\mathrm{cosec}}^{2}\text{A}-\text{2}\mathrm{cosec}\text{A}-2\mathrm{cotA}+2\mathrm{cotAcosecA}}{{\mathrm{cot}}^{2}\text{A}-1-{\mathrm{cosec}}^{2}\text{\hspace{0.17em}A}+2\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{A}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2{\mathrm{cosec}}^{2}\text{A}-\text{2}\mathrm{cosec}\text{A}-2\mathrm{cotA}+2\mathrm{cotAcosecA}}{-1-1+2\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{A}}\\ \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}}=\frac{\mathrm{cosec}\text{A(}\mathrm{cosec}\text{A}-\text{1}\right)+\mathrm{cotA}\left(\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{A}-1\right)}{\mathrm{cosec}\text{\hspace{0.17em}A}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cosec}\text{A}+\text{cot\hspace{0.17em}A}\\ \text{(vi)}\\ \text{LHS}=\sqrt{\frac{1+\mathrm{sin}\text{A}}{1-\mathrm{sin}\text{A}}}=\sqrt{\frac{1+\mathrm{sin}\text{A}}{1-\mathrm{sin}\text{A}}×\frac{1+\mathrm{sin}\text{A}}{1+\mathrm{sin}\text{A}}}\\ \text{}=\left(1+\mathrm{sin}\text{A}\right)\sqrt{\frac{1}{1-{\mathrm{sin}}^{2}\text{A}}}=\frac{1+\mathrm{sin}\text{A}}{\mathrm{cos}\text{A}}=\mathrm{tanA}+\mathrm{secA}=\text{RHS}\\ \text{(vii)}\\ \text{LHS}=\frac{\text{sin}\mathrm{\theta }-{\text{2sin}}^{3}\mathrm{\theta }}{{\text{2cos}}^{3}\mathrm{\theta }-\text{cos}\mathrm{\theta }}\\ \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}}=\frac{\text{sin}\mathrm{\theta }\left(1-{\text{2sin}}^{2}\mathrm{\theta }\right)}{\text{cos}\mathrm{\theta }\left({\text{2cos}}^{2}\mathrm{\theta }-1\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}}=\frac{\text{sin}\mathrm{\theta }\left(1-{\text{2sin}}^{2}\mathrm{\theta }\right)}{\text{cos}\mathrm{\theta }\left(\text{2(1}-{\mathrm{sin}}^{2}\mathrm{\theta }\right)-1\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}}=\frac{\text{sin}\mathrm{\theta }\left(1-{\text{2sin}}^{2}\mathrm{\theta }\right)}{\text{cos}\mathrm{\theta }\left(\text{2}-2{\mathrm{sin}}^{2}\mathrm{\theta }-1\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}}=\frac{\text{sin}\mathrm{\theta }\left(1-{\text{2sin}}^{2}\mathrm{\theta }\right)}{\text{cos}\mathrm{\theta }\left(1-2{\mathrm{sin}}^{2}\mathrm{\theta }\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}}=\mathrm{tan\theta }=\mathrm{RHS}\\ \text{(viii)}\\ \text{LHS}={\left(\text{sin A}+\text{cosec A}\right)}^{2}+{\left(\text{cos A}+\text{sec A}\right)}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{sin}}^{2}\mathrm{A}+{\mathrm{cosec}}^{2}\mathrm{A}+2+{\mathrm{cos}}^{2}\mathrm{A}+{\mathrm{sec}}^{2}\mathrm{A}+2\\ \text{}={\mathrm{sin}}^{2}\mathrm{A}+{\mathrm{cos}}^{2}\mathrm{A}+4+{\mathrm{sec}}^{2}\mathrm{A}+{\mathrm{cosec}}^{2}\mathrm{A}\\ \text{}=1+4+1+{\mathrm{tan}}^{2}\mathrm{A}+1+{\mathrm{cot}}^{2}\mathrm{A}\\ \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}}=7+{\mathrm{tan}}^{2}\mathrm{A}+{\mathrm{cot}}^{2}\mathrm{A}=\text{RHS}\\ \text{(ix)}\\ \text{LHS}=\text{(cosec A}-\text{sin A) (sec A}-\text{cos A)}\\ \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}}=\left(\frac{1}{\mathrm{sinA}}-\mathrm{sinA}\right)\left(\frac{1}{\mathrm{cosA}}-\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}}=\left(\frac{1-{\mathrm{sin}}^{2}\mathrm{A}}{\mathrm{sinA}}\right)\left(\frac{1-{\mathrm{cos}}^{2}\mathrm{A}}{\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}}=\frac{{\mathrm{cos}}^{2}\mathrm{A}}{\mathrm{sinA}}×\frac{{\mathrm{sin}}^{2}\mathrm{A}}{\mathrm{cosA}}=\frac{\mathrm{sinAcosA}}{1}=\frac{\mathrm{sinAcosA}}{{\mathrm{sin}}^{2}\mathrm{A}+{\mathrm{cos}}^{2}\mathrm{A}}\\ \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}}=\frac{1}{\frac{{\mathrm{sin}}^{2}\mathrm{A}+{\mathrm{cos}}^{2}\mathrm{A}}{\mathrm{sinAcosA}}}=\frac{1}{\frac{\mathrm{sinA}}{\mathrm{cosA}}+\frac{\mathrm{cosA}}{\mathrm{sinA}}}\\ \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}{\mathrm{tanA}+\mathrm{cotA}}\\ \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{RHS}\\ \text{(x)}\\ \text{LHS}=\frac{1+{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{cot}}^{2}\text{A}}=\frac{{\mathrm{sec}}^{2}\mathrm{A}}{{\mathrm{cosec}}^{2}\mathrm{A}}=\frac{\frac{1}{{\mathrm{cos}}^{2}\mathrm{A}}}{\frac{1}{{\mathrm{sin}}^{2}\mathrm{A}}}=\frac{{\mathrm{sin}}^{2}\mathrm{A}}{{\mathrm{cos}}^{2}\mathrm{A}}={\mathrm{tan}}^{2}\mathrm{A}=\mathrm{RHS}\\ \text{LHS}={\left(\frac{\text{1}-\text{tanA}}{\text{1}-\text{cot A}}\right)}^{2}={\left(\frac{1-\frac{\mathrm{sinA}}{\mathrm{cosA}}}{1-\frac{\mathrm{cosA}}{\mathrm{sinA}}}\right)}^{2}={\left(\frac{\frac{\mathrm{cosA}-\mathrm{sinA}}{\mathrm{cosA}}}{\frac{\mathrm{sinA}-\mathrm{cosA}}{\mathrm{sinA}}}\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}\hspace{0.17em}}=\frac{{\mathrm{sin}}^{2}\mathrm{A}}{{\mathrm{cos}}^{2}\mathrm{A}}={\mathrm{tan}}^{2}\mathrm{A}=\mathrm{RHS}\end{array}$

## NCERT Solutions for Class 10 Maths Related Chapters

### 1. Where will I find the NCERT Class 10 Mathematics Chapter 8 Solutions?

You will find the NCERT Class 10 Mathematics Chapter 8 Solutions on Extramarks official website.

### 2. How to prepare for the Mathematics examination with the help of NCERT Solutions?

The steps to prepare for the examination with the help of  NCERT Solutions are:

1. Learn the concepts
2. Revise all the formulae on a regular basis
3. Practise the exercises from every topic regularly
4. Revise all the topics and don’t skip any exercise.

### 3. What is included in Exercise 8.4 of Chapter 8 Trigonometry in Class 10 Mathematics?

The Exercise 8.4 has questions related to trigonometric ratios of complementary angles. Students will have to memorise some standard formulae in order to solve every question from the exercise.