# NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 (Ex 2.2)

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## NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 (Ex 2.2)

Mathematics Chapter-2 Class-12 is about inverse trigonometric functions. Inverse trigonometric functions are defined as the inverse functions of the basic trigonometric functions that are sine, cosine, tangent, cotangent, secant, and cosecant functions. Inverse trigonometric functions are also called arc functions as, for a given value of trigonometric functions, they produce the length of arc required to obtain that particular value. Inverse trigonometric functions perform the opposite operation of the trigonometric functions. Students have learned the basics of trigonometric functions in Class 10 and 11, which will help them build a foundation for the topic in Class 12. Inverse trigonometric functions are specially used for the problems of right-angled triangles. The inverse of these trigonometric functions is used to calculate an angle from its trigonometric ratios. The inverse trigonometric functions are also useful when solving some problems of physics. For the students to have a deep understanding and clear concepts of Class 12 Maths Chapter 2 Exercise 2.2, Extramarks provide the students with the NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2.

## Topics Covered in the NCERT Solutions For Class 12 Maths Chapter 2

Class 12 Maths Chapter 2 Exercise 2.2 Solutions contain the questions related to the properties of inverse trigonometric functions. Overall, there are 21 questions in the exercise based on the same concept. With the help of the solved examples given in the NCERT textbook, students can have a better understanding of the concepts of the Class 12 Maths NCERT Solutions Chapter 2 Exercise 2.2. Students can easily find the NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 on the internet, but the solutions provided by Extramarks are curated by the subject experts and are completely reliable. Extramarks is a one-stop solution to all the problems that students face. The NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 provided by Extramarks are detailed stepwise solutions that can be easily understood by the students as they are written in easy language. Students can refer to the authentic NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 rather than wasting their time by finding and cross-checking these answers from various other portals. Extramarks provides the students with NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 in PDF Format, therefore they are easily accessible on any device and students can go through them anytime, and anywhere they want. Also, as these solutions can be downloaded, they will be very helpful for the students to clarify their doubts.

Trigonometry has a wide range of practical applications, such as in Biology, Medical Imaging (CT scan, ultrasound), Chemistry, Number Theory, Cryptology, Metrology, Oceanology, Image Compression, Phonetics, Economics, Engineering (Electrical, Mechanical, Civil), Computer Graphics, Cartography, Crystallography and Game Development. Inverse trigonometric functions are a very essential topic for the students of Class 12. Extramarks recommends the students go through the NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 so that they can have better conceptual clarity of the topic and can score better marks. Along with the NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 Extramarks also provides students with the NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7, NCERT Solutions Class 6, NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2, NCERT Solutions Class 1 and much more so that the students can have ready access to the entire study material. Click on the given link to download the NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2.

## Access NCERT Solutions for Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions

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### Important Formulas/Properties covered in the NCERT Solutions for Class 12 Maths Chapter 2

Trigonometry is a branch of Mathematics that deals with the relationship between the sides and the angles of right-angled triangles. It involves various trigonometric functions that are used for the determination of unknown angles and sides of a triangle. The NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 provided by Extramarks help students to have deep understanding of this topic with clear basic concepts. Also, to gain a better hold of the NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2, students are required to thoroughly practice the NCERT textbook questions as well as the extra questions on the topics. Students can refer to the Extramarks’ website for the NCERT textbook solutions, as well as other practice assignments so that they can score good marks in the board examinations. Having access to the accurate, detailed and well-explained NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 is very essential for students to prepare for their board examinations.

### Complimentary Functions Related to Inverse Trigonometric Function

The trigonometric Function is a concept that has a wide range of applications, that is why it is included in the curriculum of Class 12 Mathematics. The knowledge of trigonometry is also very essential for some subjects taught in higher-level studies. Therefore, Extramarks provides students with the NCERT Textbook solutions and handy notes of Chapter 2, Class 12 Mathematics like the NCERT Solutions Class 12 Maths Chapter 2 Exercise 2.2.

Trigonometry has many geometric and analytical applications. This concept is widely used in the fields of Engineering like Electrical and Mechanical Engineering. Also, trigonometry has been included in the curriculum of Class 10 Mathematics because it is very essential for the students to have a basic understanding of the concept. It is also essential for the students to have knowledge of the NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 for building the fundamentals of the further chapters of Class 12 Mathematics.

1. sin -1 1/x = cosec-1 x, x ≥ 1 or x ≤ – 1
2. cos-1 1/x = sec-1 x, x ≥ 1 or x ≤ – 1
3. tan-1 1/x = cot-1 x, x > 0

For more information about the complementary functions related to inverse trigonometric function, students can download the NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 and subscribe to the Extramarks website.

Click on the given link to download the NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2.

### NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2

There are numerous benefits of the NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 provided by the Extramarks website. Scoring good in the Class 12 Maths board examination is not easy. Students require rigorous practice to have the curriculum of Class 12 Mathematics on their tips. The primary and the most important step in obtaining good scores in Class 12 Mathematics is to practice the NCERT Textbook. But, the NCERT Textbook just contains the answers to the problems, and just answers are not enough. Having a clear understanding of the concepts is very crucial. Therefore, Extramarks provides students with the  NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 so that they can have study material, and they can practice, learn and excel in their studies.

### NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2

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### Access Other Exercises of Class 12 Maths Chapter 2

 Chapter 2 Inverse Trigonometric Functions Exercises Exercise 2.1 14 Questions & Solutions (14 Short Answers)

Q.1 Prove the following:

${\text{3sin}}^{–1}\mathrm{x}={\mathrm{sin}}^{–1}\left(3\mathrm{x}–4{\mathrm{x}}^{3}\right),\mathrm{x}\in \left[-\frac{1}{2},\frac{1}{2}\right]$

Ans

$\begin{array}{l}\mathrm{R}.\mathrm{H}.\mathrm{S}.={\mathrm{sin}}^{-1}\left(3\mathrm{x}-4{\mathrm{x}}^{3}\right)\\ \mathrm{ }={\mathrm{sin}}^{-1}\left(3\mathrm{sin\theta }-4{\mathrm{sin}}^{3}\mathrm{\theta }\right)\left[\mathrm{Putting}\mathrm{x}=\mathrm{sin\theta }\right]\\ \mathrm{ }={\mathrm{sin}}^{-1}\left(\mathrm{sin}3\mathrm{\theta }\right)\left[\because \mathrm{sin}3\mathrm{x}=3\mathrm{sinx}-4{\mathrm{sin}}^{3}\mathrm{x}\right]\\ \mathrm{ }=3\mathrm{\theta }\\ \mathrm{ }=3{\mathrm{sin}}^{-1}\mathrm{x}\left[\therefore \mathrm{x}=\mathrm{sin\theta }⇒\mathrm{\theta }={\mathrm{sin}}^{-1}\mathrm{x}\right]\\ \mathrm{ }=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.2

${\text{3cos}}^{–1}x=co{s}^{–1}\left(4{x}^{3}-3x\right),x\in \left[\frac{1}{2},1\right]$

Ans

$\begin{array}{l}\mathrm{R}.\mathrm{H}.\mathrm{S}.={\mathrm{cos}}^{-1}\left(4{\mathrm{x}}^{3}-3\mathrm{x}\right)\\ ={\mathrm{cos}}^{-1}\left(4{\mathrm{cos}}^{3}\mathrm{\theta }-3\mathrm{cos\theta }\right)\left[\mathrm{Putting}\mathrm{x}=\mathrm{cos\theta }\right]\\ ={\mathrm{cos}}^{-1}\left(\mathrm{cos}3\mathrm{\theta }\right)\left[\because \mathrm{cos}3\mathrm{x}=4{\mathrm{cos}}^{3}\mathrm{x}-3\mathrm{cosx}\right]\\ =3\mathrm{\theta }\\ =3{\mathrm{cos}}^{-1}\mathrm{x}\left[\therefore \mathrm{x}=\mathrm{cos\theta }⇒\mathrm{\theta }={\mathrm{cos}}^{-1}\mathrm{x}\right]\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.3

${\text{tan}}^{–1}\left(\frac{2}{11}\right)+{\mathrm{tan}}^{–1}\left(\frac{7}{24}\right)={\mathrm{tan}}^{–1}\left(\frac{1}{2}\right)$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.={\mathrm{tan}}^{–1}\left(\frac{\mathrm{2}}{\mathrm{11}}\right)+{\mathrm{tan}}^{–1}\left(\frac{\mathrm{7}}{\mathrm{24}}\right)\\ ={\mathrm{tan}}^{–1}\mathrm{}\left(\frac{\frac{\mathrm{2}}{\mathrm{11}}\mathrm{+}\frac{\mathrm{7}}{\mathrm{24}}}{1-\frac{2}{11}×\frac{7}{24}}\right)\mathrm{}\left[\because {\mathrm{tan}}^{-1}\mathrm{x}+{\mathrm{tan}}^{-1}\mathrm{y}={\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)\right]\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{48+77}{11×24}}{\frac{11×24-14}{11×24}}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{125}{250}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{1}{2}\right) =\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.4

${\text{2tan}}^{–1}\left(\frac{1}{2}\right)+{\mathrm{tan}}^{–1}\left(\frac{1}{7}\right)={\mathrm{tan}}^{–1}\left(\frac{31}{17}\right)$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}=2{\mathrm{tan}}^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{\mathrm{tan}}^{–1}\left(\frac{\mathrm{1}}{\mathrm{7}}\right)\\ \mathrm{ }={\mathrm{tan}}^{-1}\left(\frac{2×\frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^{2}}\right)+{\mathrm{tan}}^{–1}\left(\frac{\mathrm{1}}{\mathrm{7}}\right)\left[\because 2{\mathrm{tan}}^{-1}\mathrm{x}={\mathrm{tan}}^{-1}\left(\frac{2\mathrm{x}}{1-{\mathrm{x}}^{2}}\right)\right]\\ \mathrm{ }={\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)+{\mathrm{tan}}^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{7}}\right)\\ \mathrm{ }={\mathrm{tan}}^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}×\frac{1}{7}}\right)\left[\because {\mathrm{tan}}^{-1}\mathrm{x}+{\mathrm{tan}}^{-1}\mathrm{y}={\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)\right]\\ \mathrm{ }={\mathrm{tan}}^{–1}\mathrm{}\left(\frac{\mathrm{31}}{\mathrm{17}}\right)\\ \mathrm{ }=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.5 Write the following functions in the simplest form:

${\text{tan}}^{–1}\frac{\sqrt{1+{\mathrm{x}}^{2}}–1}{\mathrm{x}},\mathrm{x}\ne 0$

Ans

$\begin{array}{l}\mathrm{Given}:\\ {\mathrm{tan}}^{-1}\frac{\sqrt{1+{\mathrm{x}}^{2}}-1}{\mathrm{x}}={\mathrm{tan}}^{-1}\frac{\sqrt{1+{\mathrm{tan}}^{2}\mathrm{\theta }}-1}{\mathrm{tan\theta }}\mathrm{ }\left[\mathrm{Putting}\mathrm{x}=\mathrm{tan\theta }\right]\\ \mathrm{ }={\mathrm{tan}}^{-1}\frac{\sqrt{{\mathrm{sec}}^{2}\mathrm{\theta }}-1}{\mathrm{tan\theta }}\left[\because 1+{\mathrm{tan}}^{2}\mathrm{\theta }={\mathrm{sec}}^{2}\mathrm{\theta }\right]\\ \mathrm{ }={\mathrm{tan}}^{-1}\frac{\mathrm{sec\theta }-1}{\mathrm{tan\theta }}\\ \mathrm{ }={\mathrm{tan}}^{-1}\frac{1-\mathrm{cos\theta }}{\mathrm{sin\theta }}\\ \mathrm{ }={\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{\mathrm{\theta }}{2}\right)\right)\left[\because {\mathrm{tan}}^{-1}\left(\mathrm{tanx}\right)=\mathrm{x}\right]\\ \mathrm{ }=\frac{\mathrm{\theta }}{2}\\ \mathrm{ }=\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}\end{array}$

Q.6

${\text{tan}}^{–1}\frac{1}{\sqrt{{\mathrm{x}}^{2}-1}},|\mathrm{x}|>1$

Ans

$\begin{array}{l}\mathrm{Given}:\\ {\mathrm{tan}}^{-1}\frac{1}{\sqrt{{\mathrm{x}}^{2}-1}}={\mathrm{tan}}^{-1}\frac{1}{\sqrt{{\mathrm{sec}}^{\mathrm{2}}\mathrm{\theta }-1}}\mathrm{ }\left[\mathrm{Putting}\mathrm{x}=\mathrm{sec\theta }\right]\\ \mathrm{ }={\mathrm{tan}}^{-1}\frac{1}{\mathrm{tan\theta }}\left[\because 1+{\mathrm{tan}}^{2}\mathrm{\theta }={\mathrm{sec}}^{2}\mathrm{\theta }\right]\\ \mathrm{ }={\mathrm{tan}}^{-1}\left(\mathrm{cot\theta }\right)\\ \mathrm{ }={\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)\right)\\ \mathrm{ }=\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\left[\because {\mathrm{tan}}^{-1}\left(\mathrm{tanx}\right)=\mathrm{x}\right]\\ \mathrm{ }=\frac{\mathrm{\pi }}{2}-{\mathrm{sec}}^{-1}\mathrm{x}\left[\because \mathrm{x}=\mathrm{sec\theta }\right]\\ \mathrm{ }={\mathrm{cosec}}^{-1}\mathrm{x}\end{array}$

Q.7

$ta{n}^{–1}\left(\sqrt{\frac{1–cosx}{1+cosx}}\right),\text{}\text{}x<\pi$

Ans

$\begin{array}{l}{\text{tan}}^{\text{-1}}\left(\sqrt{\frac{\text{1-cosx}}{\text{1+cosx}}}\right)={\text{tan}}^{\text{-1}}\left(\sqrt{\frac{{\text{2sin}}^{\text{2}}\left(\frac{\text{x}}{2}\right)}{{\text{2cos}}^{\text{2}}\left(\frac{x}{2}\right)}}\right)\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}\because \mathrm{cos}2x=1-2{\mathrm{sin}}^{2}x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2{\mathrm{cos}}^{2}x-1\end{array}\right]\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{x}{2}\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{x}{2}\text{}\text{}\text{}\text{}\text{}\left[\because {\mathrm{tan}}^{-1}\left(\mathrm{tan}x\right)=x\right]\end{array}$

Q.8

${\text{tan}}^{–1}\left(\frac{\mathrm{cosx}–\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}}\right),0<\mathrm{x}<\mathrm{\pi }$

Ans

$\begin{array}{l}{\mathrm{tan}}^{–1}\left(\frac{\mathrm{cosx}-\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}}\right)={\mathrm{tan}}^{-1}\left(\frac{1-\mathrm{tanx}}{1+\mathrm{tanx}}\right)\left[\begin{array}{l}\mathrm{Dividing}\mathrm{numerator}\\ \mathrm{and}\mathrm{denominator}\mathrm{by}\mathrm{sinx}\end{array}\right]\\ ={\mathrm{tan}}^{-1}\left\{\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)\right\}\left[\because \mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)=\frac{1-\mathrm{tanx}}{1+\mathrm{tanx}}\right]\\ =\frac{\mathrm{\pi }}{4}-\mathrm{x}\left[\because {\mathrm{tan}}^{-1}\left(\mathrm{tanx}\right)=\mathrm{x}\right]\end{array}$

Q.9

${\text{tan}}^{–1}\frac{\mathrm{x}}{\sqrt{{\mathrm{a}}^{2}-{\mathrm{x}}^{2}}},|\mathrm{x}|<\mathrm{a}$

Ans

$\begin{array}{l}{\mathrm{tan}}^{–1}\frac{\mathrm{x}}{\sqrt{{\mathrm{a}}^{\mathrm{2}}–{\mathrm{x}}^{\mathrm{2}}}}={\mathrm{tan}}^{–1}\frac{\mathrm{asin\theta }}{\sqrt{{\mathrm{a}}^{\mathrm{2}}-{\mathrm{a}}^{\mathrm{2}}{\mathrm{sin}}^{2}\mathrm{\theta }}}\left[\mathrm{Putting}\mathrm{x}=\mathrm{a}\mathrm{sin}\mathrm{\theta }\right]\\ \mathrm{ }={\mathrm{tan}}^{–1}\mathrm{}\left(\frac{\mathrm{asin\theta }}{\mathrm{acos\theta }}\right)\left[\because 1-{\mathrm{sin}}^{2}\mathrm{\theta }={\mathrm{cos}}^{2}\mathrm{\theta }\right]\\ \mathrm{ }={\mathrm{tan}}^{–1}\left(\mathrm{tan\theta }\right)\left[\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}=\mathrm{tan\theta }\right]\\ \mathrm{ }=\mathrm{\theta }\\ \mathrm{ }={\mathrm{sin}}^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\left[\because \mathrm{x}=\mathrm{asin\theta }\right]\end{array}$

Q.10

${\text{tan}}^{–1}\left(\frac{3{\mathrm{a}}^{2}\mathrm{x}–{\mathrm{x}}^{3}}{{\mathrm{a}}^{3}–3{\mathrm{ax}}^{2}}\right),\mathrm{a}>0;-\frac{\mathrm{a}}{\sqrt{3}}\le \mathrm{x}\le \frac{\mathrm{a}}{\sqrt{3}}$

Ans

$\begin{array}{l}{\mathrm{tan}}^{–1}\mathrm{}\left(\frac{3{\mathrm{a}}^{\mathrm{2}}\mathrm{x}-{\mathrm{x}}^{\mathrm{3}}}{{\mathrm{a}}^{\mathrm{3}}-3{\mathrm{ax}}^{\mathrm{2}}}\right)={\mathrm{tan}}^{–1}\mathrm{}\left(\frac{3{\mathrm{a}}^{\mathrm{3}}\mathrm{tan\theta }-{\left(\mathrm{atan\theta }\right)}^{\mathrm{3}}}{{\mathrm{a}}^{\mathrm{3}}-3\mathrm{a}{\left(\mathrm{atan\theta }\right)}^{\mathrm{2}}}\right)\left[\mathrm{Putting}\mathrm{x}=\mathrm{atan\theta }\right]\\ ={\mathrm{tan}}^{–1}\left(\frac{3{\mathrm{a}}^{\mathrm{3}}\mathrm{tan\theta }-{\mathrm{a}}^{3}{\mathrm{tan}}^{3}\mathrm{\theta }}{{\mathrm{a}}^{\mathrm{3}}-3{\mathrm{a}}^{\mathrm{3}}{\mathrm{tan}}^{2}\mathrm{\theta }}\right)\\ ={\mathrm{tan}}^{–1}\mathrm{}\left\{\frac{\overline{){\mathrm{a}}^{3}}\left(\mathrm{3}\mathrm{tan\theta }-{\mathrm{tan}}^{3}\mathrm{\theta }\right)}{\overline{){\mathrm{a}}^{\mathrm{3}}}\left(\mathrm{1}-\mathrm{3}{\mathrm{tan}}^{2}\mathrm{\theta }\right)}\right\}\\ ={\mathrm{tan}}^{–1}\left(\mathrm{tan}3\mathrm{\theta }\right)\left[\mathrm{tan}3\mathrm{\theta }=\frac{3\mathrm{tan\theta }-{\mathrm{tan}}^{3}\mathrm{\theta }}{1-3{\mathrm{tan}}^{2}\mathrm{\theta }}\right]\\ =3\mathrm{\theta }\\ =3{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\left[\because \mathrm{x}=\mathrm{atan\theta }\right]\end{array}$

Q.11 Find the values of each of the following:

${\text{tan}}^{–1}\left\{2\mathrm{cos}\left(2{\mathrm{sin}}^{–1}\frac{1}{2}\right)\right\}$

Ans

$\begin{array}{l}{\mathrm{tan}}^{-1}\left\{2\mathrm{cos}\left(2{\mathrm{sin}}^{–1}\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}{\mathrm{tan}}^{-1}\mathrm{}\left\{2\mathrm{cos}\left(\mathrm{2}×\frac{\mathrm{\pi }}{6}\right)\right\}\\ ={\mathrm{tan}}^{-1}\left\{2\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)\right\}\\ ={\mathrm{tan}}^{-1}\left\{\mathrm{2}×\frac{1}{2}\right\}\\ ={\mathrm{tan}}^{-1}\left(1\right)\mathrm{ }=\frac{\mathrm{\pi }}{4}\end{array}$

Q.12

$\text{cot}\left({\mathrm{tan}}^{–1}\mathrm{a}+{\mathrm{cot}}^{–1}\mathrm{a}\right)$

Ans

$\begin{array}{l}\mathrm{cot}\mathrm{}\left({\mathrm{tan}}^{–1}\mathrm{a}+{\mathrm{cot}}^{–1}\mathrm{a}\right)=\mathrm{cot}\left[\because {\mathrm{tan}}^{–1}\mathrm{x}+{\mathrm{cot}}^{–1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\right]\mathrm{}\left(\frac{\mathrm{\pi }}{2}\right)\mathrm{}\\ =0\end{array}$

Q.13

$\text{tan}\frac{1}{2}\left({\mathrm{sin}}^{–1}\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}+{\mathrm{cos}}^{–1}\frac{1–{\mathrm{y}}^{2}}{1+{\mathrm{y}}^{2}}\right),|\mathrm{x}|<1,\mathrm{y}>0\mathrm{and}\mathrm{xy}<1$

Ans

$\begin{array}{l}\mathrm{tan}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{}\left({\mathrm{sin}}^{–1}\frac{2\mathrm{x}}{\mathrm{1}+{\mathrm{x}}^{\mathrm{2}}}+{\mathrm{cos}}^{–1}\frac{\mathrm{1}-{\mathrm{y}}^{\mathrm{2}}}{\mathrm{1}+{\mathrm{y}}^{\mathrm{2}}}\right)\\ =\mathrm{tan}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{}\left({\mathrm{sin}}^{–1}\frac{\mathrm{2}\mathrm{tan\alpha }}{\mathrm{1}+{\mathrm{tan}}^{2}\mathrm{\alpha }}+{\mathrm{cos}}^{–1}\frac{\mathrm{1}-{\mathrm{tan}}^{2}\mathrm{\beta }}{\mathrm{1}+{\mathrm{tan}}^{2}\mathrm{\beta }}\right)\left[\mathrm{Putting}\mathrm{x}=\mathrm{tan\alpha },\mathrm{}\mathrm{y}=\mathrm{tan\beta }\right]\\ =\mathrm{tan}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{}\left\{{\mathrm{sin}}^{–1}\left(\mathrm{sin}2\mathrm{\alpha }\right)+{\mathrm{cos}}^{–1}\left(\mathrm{cos}2\mathrm{\beta }\right)\right\}\\ =\mathrm{tan}\frac{\mathrm{1}}{\mathrm{2}}\left(2\mathrm{\alpha }+2\mathrm{\beta }\right)\\ =\mathrm{tan}\left(\mathrm{\alpha }+\mathrm{\beta }\right)\\ =\frac{\mathrm{tan\alpha }+\mathrm{tan\beta }}{1-\mathrm{tan\alpha tan\beta }}\\ =\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\end{array}$

Q.14

$\mathrm{If}\mathrm{sin}\left({\mathrm{sin}}^{–1}\frac{1}{5}+{\mathrm{cos}}^{–1}\mathrm{x}\right)=1,\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{sin}\left({\mathrm{sin}}^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\mathrm{cos}}^{–1}\mathrm{x}\right)=\mathrm{1}\\ ⇒\mathrm{sin}\left({\mathrm{sin}}^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\mathrm{cos}}^{–1}\mathrm{x}\right)\mathrm{}={\mathrm{sin}}^{–1}\left(\frac{\mathrm{\pi }}{2}\right)\\ ⇒ \mathrm{ }{\mathrm{sin}}^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\mathrm{cos}}^{–1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\\ ⇒ \mathrm{ }{\mathrm{sin}}^{–1}\frac{\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{\pi }}{2}-{\mathrm{cos}}^{–1}\mathrm{x}\\ ⇒ \mathrm{ }{\mathrm{sin}}^{–1}\frac{\mathrm{1}}{\mathrm{5}}={\mathrm{sin}}^{–1}\mathrm{x}\left[\because {\mathrm{sin}}^{–1}\mathrm{x}+{\mathrm{cos}}^{–1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\right]\\ ⇒ \frac{1}{5}=\mathrm{x}\end{array}$

Q.15

${\text{If tan}}^{–1}\frac{\mathrm{x}–1}{\mathrm{x}–2}+{\mathrm{tan}}^{–1}\frac{\mathrm{x}+1}{\mathrm{x}+2}=\frac{\mathrm{\pi }}{4},\mathrm{}\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}.$

Ans

$\begin{array}{l} \mathrm{ }{\mathrm{tan}}^{–1}\frac{\mathrm{x}-\mathrm{1}}{\mathrm{x}-\mathrm{2}}+{\mathrm{tan}}^{–1}\frac{\mathrm{x}+1}{\mathrm{x}+2}=\frac{\mathrm{\pi }}{4}\\ ⇒{\mathrm{tan}}^{–1}\left(\frac{\frac{\mathrm{x}-\mathrm{1}}{\mathrm{x}-\mathrm{2}}+\frac{\mathrm{x}+1}{\mathrm{x}+2}}{1-\left(\frac{\mathrm{x}-\mathrm{1}}{\mathrm{x}-\mathrm{2}}\right)\left(\frac{\mathrm{x}+1}{\mathrm{x}+2}\right)}\right)=\frac{\mathrm{\pi }}{4}\left[\because {\mathrm{tan}}^{-1}\mathrm{x}+{\mathrm{tan}}^{-1}\mathrm{y}={\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)\right]\\ ⇒ \mathrm{ }\left(\frac{\frac{{\mathrm{x}}^{\mathrm{2}}+\mathrm{x}-2+{\mathrm{x}}^{\mathrm{2}}-\mathrm{x}-\mathrm{2}}{\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}}{\frac{{\mathrm{x}}^{\mathrm{2}}-\mathrm{4}-{\mathrm{x}}^{\mathrm{2}}+1}{\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}}\right)=\mathrm{tan}\frac{\mathrm{\pi }}{4}\\ ⇒ \mathrm{ }\left(\frac{2{\mathrm{x}}^{2}-4}{-3}\right)=1\\ ⇒ 2{\mathrm{x}}^{2}-4=-3\\ ⇒ \mathrm{x}=±\frac{1}{\sqrt{2}}\end{array}$

Q.16

${\text{sin}}^{–1}\left(\mathrm{sin}\frac{2\mathrm{\pi }}{3}\right)$

Ans

$\begin{array}{l}\mathrm{ }{\mathrm{sin}}^{–1}\mathrm{}\left(\mathrm{sin}\frac{\mathrm{2}\mathrm{\pi }}{\mathrm{3}}\right)\mathrm{}=\mathrm{sin}\because \mathrm{Range}\mathrm{of}\mathrm{sin\theta }\mathrm{is} {\left[-\frac{\mathrm{\pi }}{2}\mathrm{,}\frac{\mathrm{\pi }}{2}\mathrm{}\right]}^{–1}\mathrm{}\left\{\mathrm{sin}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{\mathrm{3}}\right)\right\}\\ ={\mathrm{sin}}^{–1}\left\{\mathrm{sin}\frac{\mathrm{\pi }}{\mathrm{3}}\right\}\left[\because \mathrm{sin}\left(\mathrm{\pi }-\mathrm{\theta }\right)=\mathrm{sin\theta }\right]\\ =\frac{\mathrm{\pi }}{\mathrm{3}}\end{array}$

Q.17

${\text{tan}}^{–1}\left(tan\frac{3\pi }{4}\right)$

Ans

$\begin{array}{l}{\mathrm{tan}}^{–1}\left(\mathrm{tan}\frac{\mathrm{3}\mathrm{\pi }}{\mathrm{4}}\right)={\mathrm{tan}}^{–1}\left\{\mathrm{tan}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{\mathrm{4}}\right)\right\}\left[\because \mathrm{Range}\mathrm{of}\mathrm{tan\theta }\mathrm{is} \left(-\frac{\mathrm{\pi }}{2}\mathrm{,}\frac{\mathrm{\pi }}{2}\right)\right]\\ ={\mathrm{tan}}^{–1}\mathrm{}\left\{-\mathrm{tan}\frac{\mathrm{\pi }}{\mathrm{4}}\right\}\left[\because \mathrm{tan}\left(\mathrm{\pi }-\mathrm{\theta }\right)=-\mathrm{tan\theta }\right]\\ ={\mathrm{tan}}^{–1}\left\{\mathrm{tan}\left(-\frac{\mathrm{\pi }}{\mathrm{4}}\right)\right\}\mathrm{}\left[\mathrm{tan}\left(-\mathrm{\theta }\right)=-\mathrm{tan\theta }\right]\\ =-\frac{\mathrm{\pi }}{\mathrm{4}}\end{array}$

Q.18

$\text{tan}\left({\mathrm{sin}}^{–1}\frac{3}{5}+{\mathrm{cot}}^{–1}\frac{3}{2}\right)$

Ans

$\begin{array}{l}\mathrm{tan}\mathrm{}\left({\mathrm{sin}}^{–1}\frac{\mathrm{3}}{\mathrm{5}}+{\mathrm{cot}}^{–1}\frac{\mathrm{3}}{\mathrm{2}}\right)\\ \mathrm{Let}{\mathrm{sin}}^{–1}\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{x} \mathrm{and} {\mathrm{cot}}^{–1}\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{y}\\ ⇒ \mathrm{ }\mathrm{sinx}=\frac{3}{5} \mathrm{and} \mathrm{coty}=\frac{3}{2}\\ ⇒\mathrm{cosx}=\sqrt{1-{\mathrm{sin}}^{2}\mathrm{x}}\\ =\sqrt{1-{\left(\frac{3}{5}\right)}^{2}}=\frac{4}{5}\\ \mathrm{and} \mathrm{ }\mathrm{tan}\mathrm{}\mathrm{y}=\frac{1}{\mathrm{cot}\mathrm{y}}\\ \mathrm{ }=\frac{1}{\left(\frac{3}{2}\right)}=\frac{2}{3}\\ ⇒\mathrm{tanx}=\frac{\mathrm{sinx}}{\mathrm{cosx}}=\frac{\left(\frac{3}{5}\right)}{\left(\frac{4}{5}\right)}\\ =\frac{3}{4}\\ ⇒\mathrm{x}={\mathrm{tan}}^{-1}\left(\frac{3}{4}\right) \mathrm{and}\mathrm{ }\mathrm{y}={\mathrm{tan}}^{-1}\left(\frac{2}{3}\right)\\ ⇒{\mathrm{sin}}^{–1}\frac{\mathrm{3}}{\mathrm{5}}={\mathrm{tan}}^{-1}\left(\frac{3}{4}\right) \mathrm{and} {\mathrm{cot}}^{–1}\frac{\mathrm{3}}{\mathrm{2}}={\mathrm{tan}}^{-1}\left(\frac{2}{3}\right)\\ \therefore \mathrm{tan}\left({\mathrm{sin}}^{–1}\frac{\mathrm{3}}{\mathrm{5}}+{\mathrm{cot}}^{–1}\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{tan}\left\{{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{3}{4}\right)+{\mathrm{tan}}^{-1}\left(\frac{2}{3}\right)\right\}\\ =\mathrm{tan}\left\{{\mathrm{tan}}^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}×\frac{2}{3}}\right)\right\}\\ =\mathrm{tan}\left\{{\mathrm{tan}}^{-1}\left(\frac{17}{6}\right)\right\}\\ =\frac{17}{6}\end{array}$

Q.19

$\begin{array}{l}{\mathrm{cos}}^{–1}\left(\mathrm{cos}\frac{7\mathrm{\pi }}{6}\right)\mathrm{}\mathrm{is}\mathrm{equal}\mathrm{to}\\ \left(\mathrm{A}\right)\frac{7\mathrm{\pi }}{6}\\ \left(\mathrm{B}\right)\frac{5\mathrm{\pi }}{6}\\ \left(\mathrm{C}\right)\frac{\mathrm{\pi }}{3}\\ \left(\mathrm{D}\right)\frac{\mathrm{\pi }}{6}\end{array}$

Ans

$\begin{array}{l}{\mathrm{cos}}^{–1}\mathrm{}\left(\mathrm{cos}\frac{\mathrm{7}\mathrm{\pi }}{\mathrm{6}}\right)={\mathrm{cos}}^{–1}\left\{\mathrm{cos}\left(2\mathrm{\pi }-\frac{5\mathrm{\pi }}{\mathrm{6}}\right)\right\}\left[\because \mathrm{Range}\mathrm{of}\mathrm{cos\theta }\mathrm{is}\mathrm{ }\left[0\mathrm{,}\mathrm{\pi }\right]\right]\\ ={\mathrm{cos}}^{–1}\mathrm{}\left\{\mathrm{cos}\left(\frac{5\mathrm{\pi }}{\mathrm{6}}\right)\right\}\left[\because \mathrm{cos}\left(2\mathrm{\pi }-\mathrm{\theta }\right)=\mathrm{cos\theta }\right]\\ =\frac{5\mathrm{\pi }}{\mathrm{6}}\left[\because {\mathrm{cos}}^{-1}\left(\mathrm{cosx}\right)=\mathrm{x}\right]\\ \mathrm{Hence}\mathrm{option}\left(\mathrm{b}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.20

$\begin{array}{l}\mathrm{sin}\left(\frac{\mathrm{\pi }}{3}-{\mathrm{sin}}^{–1}\left(-\frac{1}{2}\right)\right)\mathrm{}\mathrm{is}\mathrm{equal}\mathrm{to}\\ \left(\mathrm{A}\right)\frac{1}{2}\\ \left(\mathrm{B}\right)\frac{1}{3}\\ \left(\mathrm{C}\right)\frac{1}{4}\\ \left(\mathrm{D}\right)1\end{array}$

Ans

$\begin{array}{l}\mathrm{sin}\left(\frac{\mathrm{\pi }}{\mathrm{3}}-{\mathrm{sin}}^{–1}\left(-\frac{\mathrm{1}}{\mathrm{2}}\right)\right)=\mathrm{sin}\mathrm{}\left(\frac{\mathrm{\pi }}{\mathrm{3}}+{\mathrm{sin}}^{–1}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\left[\mathrm{sin}\left(-\mathrm{\theta }\right)=-\mathrm{sin\theta }\right]\\ =\mathrm{sin}\mathrm{}\left(\frac{\mathrm{\pi }}{\mathrm{3}}+\frac{\mathrm{\pi }}{\mathrm{6}}\right)\left[\because \mathrm{sin}\frac{\mathrm{\pi }}{6}=\frac{1}{2}\right]\\ =\mathrm{sin}\mathrm{}\left(\frac{2\mathrm{\pi }+\mathrm{\pi }}{\mathrm{6}}\right)\\ =\mathrm{sin}\left(\mathrm{}\frac{3\mathrm{\pi }}{\mathrm{6}}\right)\\ =\mathrm{sin}\mathrm{}\left(\frac{\mathrm{\pi }}{\mathrm{2}}\right)\\ =1\left[\because \mathrm{sin}\frac{\mathrm{\pi }}{2}=1\right]\\ \mathrm{Hence},\mathrm{option}\left(\mathrm{D}\right) \mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.21

$\begin{array}{l}{\mathrm{tan}}^{-1}\sqrt{\mathrm{3}}-{\mathrm{cot}}^{-1}\left(\mathrm{–}\sqrt{\mathrm{3}}\right)\mathrm{is}\mathrm{ }\mathrm{equal}\mathrm{ }\mathrm{to}\\ \begin{array}{l}\left(\mathrm{A}\right) \mathrm{\pi }\\ \left(\mathrm{B}\right)\mathrm{ }-\frac{\mathrm{\pi }}{\mathrm{2}}\\ \left(\mathrm{C}\right)\mathrm{ }0\\ \left(\mathrm{D}\right)\mathrm{ }2\sqrt{\mathrm{3}}\end{array}\end{array}$

Ans

$\begin{array}{l}{\mathrm{tan}}^{–1}\sqrt{\mathrm{3}}-{\mathrm{cot}}^{–1}\left(-\sqrt{\mathrm{3}}\right)\\ =-\mathrm{ }{\mathrm{tan}}^{–1}\left(-\sqrt{\mathrm{3}}\right)-{\mathrm{cot}}^{–1}\left(-\sqrt{\mathrm{3}}\right)\left[\because \mathrm{tan}\left(-\mathrm{\theta }\right)=\mathrm{tan\theta }\right]\\ =-\mathrm{ }\left\{{\mathrm{tan}}^{–1}\left(-\sqrt{\mathrm{3}}\right)+{\mathrm{cot}}^{–1}\left(-\sqrt{\mathrm{3}}\right)\right\}\\ =-\frac{\mathrm{\pi }}{2}\left[\because {\mathrm{tan}}^{-1}\mathrm{x}+{\mathrm{con}}^{-1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\right]\\ \mathrm{Hence}\mathrm{option}\left(\mathrm{B}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$