# NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

## NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions: Two chapters 'relation and function' and 'inverse trigonometry' of NCERT Class 12 has 10 % weightage in the board examination. In this article, you will find NCERT solutions for class 12 maths chapter 2 inverse trigonometric functions which will help you to understand the concepts in a detailed manner. In class 11 maths you have already learnt about trigonometric functions. It won't take much effort to command on inverse trigonometric functions if you have good knowledge of trigonometric functions. You just need to practice NCERT questions including examples and miscellaneous exercise. You may find some difficulties in solving the problems, so you can take the help of these solutions of NCERT for class 12 maths chapter 2 inverse trigonometric functions. These solutions of NCERT for class 12 maths chapter 2 inverse trigonometric functions are very important for the board exam and as well as for competitive exams like JEE Main, BITSAT, VITEEE, etc. In this chapter, there are 2 exercises with 35 questions. All these questions are prepared and explained in a step-by-step method in the NCERT solutions for class 12 maths chapter 2 inverse trigonometric functions article.  So, it will be very easy for you to understand the concept. Check all  NCERT solutions from class 6 to class 12 to get a better understanding of the concepts.

There are important applications of ITF in geometry, navigation, science, and engineering. Also, inverse trigonometric functions play an important role in the calculus part of mathematics to define many integrals. Many students have a misconception in class 12 maths chapter inverse trigonometric functions like$\sin^-^1x=\frac{1}{\sin x}$  . But the inverse function ($f^-^1$) is not the same as  $\frac{1}{f}$ for example $\sin^-^1x\neq \frac{1}{\sin x}$.

For inverse to exist, the function must be one-one and onto but trigonometric functions are neither one-one and onto over their domain and natural ranges. So, to ensure the existence of their inverse we restrict domains and ranges of trigonometric functions. And this range is known as principal value. In the following table the principal value branches of inverse trigonometric functions(ITF) are given:

## Topics of NCERT Grade 12 Maths Chapter-2 Inverse Trigonometric Functions

2.1 Introduction

2.2 Basic Concepts

2.3 Properties of Inverse Trigonometric Functions

## CBSE NCERT Solutions for class 12 maths chapter 2 Inverse Trigonometric Functions: Exercise 2.1

Let  $x = \sin^{-1}\left ( \frac{-1}{2} \right )$

$\implies \sin x = \frac{-1}{2}= -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$
We know, principle value range of $\dpi{100} sin^{-1}$ is $\dpi{100} [-\frac{\pi}{2}, \frac{\pi}{2}]$

$\dpi{100} \therefore$ The principal value of $\dpi{100} \sin^{-1}\left ( \frac{-1}{2} \right )$ is $\dpi{100} -\frac{\pi}{6},$

So, let us assume that $\cos^{-1}\left(\frac{\sqrt3}{2} \right ) = x$  then,

Taking inverse both sides we get;

$cos\ x = (\frac{\sqrt{3}}{2})$,  or   $cos (\frac{\pi}{6}) = (\frac{\sqrt{3}}{2})$

and as we know that the principal values of $cos^{-1}$ is from [0,$\pi$],

Hence $cos\ x = (\frac{\sqrt{3}}{2})$     when x = $\frac{\pi}{6}$ .

Therefore, the principal value for $\cos^{-1}\left(\frac{\sqrt3}{2} \right )$is   $\frac{\pi}{6}$.

Let us assume that $\textup{cosec}^{-1}(2) = x$, then we have;

$Cosec\ x = 2$, or

$Cosec( \frac{\pi}{6}) = 2$  .

And we know the range of principal values is  $[\frac{-\pi}{2},\frac{\pi}{2}] - \left \{ 0 \right \}.$

Therefore the principal value of  $\textup{cosec}^{-1}(2)$ is $\frac{\pi}{6}$.

Let us assume that $\tan^{-1}(-\sqrt3) = x$, then we have;

$\tan x = (-\sqrt 3)$ or

$-\tan (\frac{\pi}{3}) = \tan \left ( \frac{-\pi}{3} \right ).$

and as we know that the principal value of $\tan^{-1}$ is $\left ( \frac{-\pi}{2}, \frac{\pi}{2} \right )$.

Hence the only principal value of $\tan^{-1}(-\sqrt3)$ when $x = \frac{-\pi}{3}$.

Let us assume that  $\cos^{-1}\left(-\frac{1}{2} \right ) =y$ then,

Easily we have; $\cos y = \left ( \frac{-1}{2} \right )$ or we can write it as:

$-\cos \left ( \frac{\pi}{3} \right ) = \cos \left ( \pi - \frac{\pi}{3} \right ) = \cos \left ( \frac{2\pi}{3} \right ).$

as we know that the range of the principal values of $\cos^{-1}$  is $\left [ 0,\pi \right ]$.

Hence $\frac{2\pi}{3}$ lies in the range it is a principal solution.

Given $\tan^{-1}(-1)$ so we can assume it to be equal to 'z';

$\tan^{-1}(-1) =z$  ,

$\tan z = -1$

or

$-\tan (\frac{\pi}{4}) = \tan(\frac{-\pi}{4})= -1$

And as we know the range of principal values of $\tan^{-1}$ from $\left ( \frac{-\pi}{2}, \frac{\pi}{2} \right )$.

As only one value z = $-\frac{\pi}{4}$ lies hence we have only one principal value that is $-\frac{\pi}{4}$.

Let us assume that $\sec^{-1}\left (\frac{2}{\sqrt3}\right) = z$  then,

we can also write it as; $\sec z = \left (\frac{2}{\sqrt3}\right)$.

Or $\sec (\frac{\pi}{6}) = \left (\frac{2}{\sqrt3}\right)$  and the principal values lies between  $\left [ 0, \pi \right ] - \left \{ \frac{\pi}{2} \right \}$.

Hence we get only one principal value of $\sec^{-1}\left (\frac{2}{\sqrt3}\right)$  i.e., $\frac{\pi}{6}$.

Let us assume that $\cot^{-1}(\sqrt3) = x$ , then we can write in other way,

$\cot x = (\sqrt3)$   or

$\cot (\frac{\pi}{6}) = (\sqrt3)$.

Hence when   $x=\frac{\pi}{6}$  we have $\cot (\frac{\pi}{6}) = (\sqrt3)$.

and the range of principal values of $\cot^{-1}$ lies in $\left ( 0, \pi \right )$.

Then the principal value of $\cot^{-1}(\sqrt3)$ is $\frac{\pi}{6}$

Let us assume $\cos^{-1}\left(-\frac{1}{\sqrt2} \right ) = x$;

Then we have $\cos x = \left ( \frac{-1}{\sqrt 2} \right )$

or

$-\cos (\frac{\pi}{4}) = \left ( \frac{-1}{\sqrt 2} \right )$  ,

$\cos (\pi - \frac{\pi}{4}) = \cos (\frac{3\pi}{4})$.

And we know the range of principal values of $\cos^{-1}$ is $[0,\pi]$.

So, the only principal value which satisfies $\cos^{-1}\left(-\frac{1}{\sqrt2} \right ) = x$ is $\frac{3\pi}{4}$.

Let us assume the value of $\textup{cosec}^{-1}(-\sqrt2) = y$ , then

we have $cosec\ y = (-\sqrt 2)$   or

$-cosec\ (\frac{\pi}{4}) = (-\sqrt 2) = cosec\ (\frac{-\pi}{4})$.

and the range of the principal values of $\textup{cosec}^{-1}$  lies between  $\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ] - \left \{ 0 \right \}$.

hence the principal value of $\textup{cosec}^{-1}(-\sqrt2)$ is $\frac{-\pi}{4}$.

To find the values first we declare each term to some constant ;

$tan^{-1}(1) = x$ , So we have $\tan x = 1$;

or $\tan (\frac{\pi}{4}) = 1$

Therefore, $x = \frac{\pi}{4}$

$cos^{-1}(\frac{-1}{2}) = y$

So, we have

$\cos y = \left ( \frac{-1}{2} \right ) = -\cos \left ( \frac{\pi}{3} \right ) = \cos(\pi - \frac{\pi}{3}) = \cos \left ( \frac{2\pi}{3} \right )$.

Therefore $y = \frac{2\pi}{3}$,

$\sin^{-1}(\frac{-1}{2}) = z$,

So we have;

$\sin z = \frac{-1}{2}$   or  $-\sin (\frac{\pi}{6}) =\sin (\frac{-\pi}{6}) = \frac{-1}{2}$

Therefore $z = -\frac{\pi}{6}$

Hence we can calculate the sum:

$= \frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}$

$=\frac{3\pi + 8\pi -2\pi}{12} = \frac{9\pi}{12}=\frac{3\pi}{4}$.

Here we have $\cos^{-1}\left(\frac{1}{2} \right ) + 2\sin^{-1}\left(\frac{1}{2} \right )$

let us assume that the value of

$\cos^{-1}\left ( \frac{1}{2} \right ) = x, \:and\:\sin^{-1}\left(\frac{1}{2} \right ) = y$;

then we have to find out the value of x +2y.

Calculation of x :

$\Rightarrow \cos^{-1}\left ( \frac{1}{2} \right ) = x$

$\Rightarrow \cos x = \frac{1}{2}$

$\Rightarrow \cos \frac{\pi}{3} = \frac{1}{2}$  ,

Hence $x = \frac{\pi}{3}$.

Calculation of y :

$\Rightarrow \sin^{-1}\left(\frac{1}{2} \right ) = y$

$\Rightarrow \sin y = \frac{1}{2}$

$\Rightarrow \sin \frac{\pi}{6} = \frac{1}{2}$  .

Hence $y = \frac{\pi}{6}$.

The required sum will be =     $\frac{\pi}{3}+2(\frac{\pi}{6}) = \frac{2\pi}{3}$.

(A)    $0\leq y \leq \pi$

(B)    $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

(C)     $0 < y < \pi$

(D)     $-\frac{\pi}{2} < y < \frac{\pi}{2}$

Given if  $\sin^{-1}x = y$  then,

As we know that the $\sin^{-1}$ can take values between $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ].$

Therefore, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$.

Hence answer choice (B) is correct.

(A)       $\pi$

(B)   $-\frac{\pi}{3}$

(C)       $\frac{\pi}{3}$

(D)    $\frac{2\pi}{3}$

Let us assume the values of $\tan^{-1}(\sqrt3)$ be 'x'  and $\sec^{-1}(-2)$  be 'y'.

Then we have;

$\tan^{-1}(\sqrt3) = x$     or   $\tan x = \sqrt 3$    or  $\tan \frac{\pi}{3} = \sqrt 3$   or

$x = \frac{\pi}{3}$

and  $\sec^{-1}(-2) = y$    or  $\sec y = -2$

or   $-\sec (\frac{\pi}{3}) =\sec ({\pi - \frac{\pi}{3}}) = \sec{\frac{2\pi}{3}}$

$y = \frac{2\pi}{3}$

also, the ranges of the principal values of $\tan^{-1}$ and  $\sec^{-1}$   are $(\frac{-\pi}{2},\frac{\pi}{2})$. and

$[0,\pi] - \left \{ \frac{\pi}{2} \right \}$   respectively.

$\therefore$ we have then;

$\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$

$= \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}$

## CBSE NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions: Exercise 2.2

Given to prove: $3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$

where,     $x\:\epsilon \left[-\frac{1}{2},\frac{1}{2} \right ]$.

Take $\theta= \sin ^{-1}x$  or   $x = \sin \theta$

Take R.H.S value

$\sin^{-1}(3x - 4x^3)$

$\sin^{-1}(3\sin \theta - 4\sin^3 \theta)$

$\sin^{-1}(\sin 3\theta)$                                        $\left ( \because 3\sin \theta - 4\sin^3 \theta \right = \sin 3 \theta )$

$3\theta$

$3\sin^{-1}x$   =   L.H.S

Given to prove $3\cos^{-1} x = \cos^{-1}(4x^3 - 3x), \;\;x\in\left[\frac{1}{2},1 \right ]$.

Take $\cos^{-1}x = \theta$  or $\cos \theta = x$;

Then we have;

R.H.S.

$\cos^{-1}(4x^3 - 3x)$

$\cos^{-1}(4\cos^3 \theta - 3\cos\theta)$      $\left [ \because 4\cos^3 \theta - 3\cos\theta = \cos3 \theta \right ]$

$\cos^{-1}(\cos3\theta)$

$3\theta$

$3\cos^{-1}x$ = L.H.S

Hence Proved.

Given to prove $\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24} = \tan^{-1}\frac{1}{2}$

We have L.H.S

$\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24}$

$=\tan^{-1}\frac{\frac{2}{11} + \frac{7}{24} }{1 - \left ( \frac{2}{11}\times\frac{7}{24} \right ) }$        $\left [ \because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x +y}{1 - xy} \right ]$

$=\tan^{-1}\frac{11\times 24 }{\frac{11\times24 -14}{11\times 24} }$

$=\tan^{-1}\frac{48 + 77}{264 -14}$

$=\tan^{-1}\left ( \frac{125}{250}\right ) = \tan^{-1}\left ( \frac{1}{2} \right )$

=   R.H.S

Hence proved.

Given to prove $2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}$

Then taking L.H.S.

We have $2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7}$

$=\tan^{-1} \frac{2.\frac{1}{2}}{1 - \left ( \frac{1}{2} \right )^2} + \tan^{-1} \frac{1}{7}$                    $\because 2\tan^{-1} x = \tan^{-1} \frac{2x}{1- x^2}$

$=\tan^{-1} \frac{1}{(\frac{3}{4})} + \tan^{-1} \frac{1}{7}$

$=\tan^{-1} \frac{4}{3} + \tan^{-1} \frac{1}{7}$

$=\tan^{-1} \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3}.\frac{1}{7}}$                                  $\left [ \because \tan^{-1}x + \tan^{-1} y = \tan^{-1} \frac{x +y}{1- xy}\right ]$

$=\tan^{-1} \left ( \frac{\frac{28+3}{21}}{\frac{21-4}{21}} \right )$

$=\tan^{-1} \frac{31}{17}$

=  R.H.S.

Hence proved.

We have $\tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x}$

Take

$\therefore$ $\tan^{-1} \frac {\sqrt{1+x^2} - 1}{x} = \tan^{-1}\frac{\sqrt{1+\tan^2 \Theta - 1}}{\tan \Theta}$

$=\tan^{-1}(\frac{sec \Theta-1}{tan \Theta}) = \tan^{-1}\left ( \frac{1-cos \Theta}{sin \Theta} \right )$

$=\tan^{-1}\left ( \frac {2sin^2\left ( \frac{\Theta}{2} \right )}{2sin\frac{\Theta}{2}cos\frac{\Theta}{2}} \right )$

$=\tan^{-1}\left ( \tan\frac{\Theta}{2} \right ) = \frac{\Theta}{2} =\frac{1}{2}\tan^{-1}x$

$=\frac{1}{2}\tan^{-1}x$  is the simplified form.

Given that $\tan^{-1} \frac{1}{\sqrt{x^2 -1}},\;\; |x| > 1$

Take $x =cosec\ \Theta$  or  $\Theta = cosec ^{-1}x$

$\therefore tan^{-1}\frac{1}{\sqrt{x^2-1}}$

$=tan^{-1} \frac{1}{\sqrt{cosec^2 \Theta -1}}$

$=tan^{-1}(\frac{1}{\cot \Theta})$

$=tan^{-1}(\tan \Theta)$  =  $\Theta$

$cosec^{-1}x$

$=\frac{\pi}{2}- \sec^{-1}x$        $[\because cosec^{-1}x + \sec^{-1}x = \frac{\pi}{2}]$

Given that $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi$

We have in inside the root the term : $\frac{1-\cos x}{1 + \cos x}$

Put $1-\cos x = 2\sin^2\frac{x}{2}$   and    $1+\cos x = 2\cos^2\frac{x}{2}$,

Then we have,

$=\tan^{-1}\left(\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}} \right )$

$=\tan^{-1}\left( \frac{\sin \frac{x}{2}}{\cos\frac{x}{2}} \right )$

$=\tan^{-1}(\tan\frac{x}{2}) = \frac{x}{2}$

Hence the simplest form is $\frac{x}{2}$

Given $\tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right )$   where $x\:\epsilon\:( \frac{-\pi}{4} < x < \frac{3\pi}{4})$

So,

$=\tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right )$

Taking $\cos x$ common from numerator and denominator.

We get:

$=\tan^{-1}\left(\frac{1 -(\frac{\sin x}{\cos x}) }{1+(\frac{\sin x}{\cos x}) } \right )$

$=\tan^{-1}\left(\frac{1 - \tan x }{1+\tan x } \right )$

$\tan^{-1}(1) - \tan^{-1}(\tan x)$         as, $\left [ \because \tan^{-1}x - \tan^{-1}y = \frac{x - y}{1 + xy} \right ]$

$\frac{\pi}{4} - x$ is the simplest form.

Given that $\tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}, \;\; |x| < a$

Take $x = a\sin \theta$   or

$\theta = \sin^{-1}\left ( \frac{x}{a} \right )$   and putting it in the equation above;

$\tan^{-1} \frac{a\sin \theta}{\sqrt{a^2 - (a\sin \theta)^2}}$

$=\tan^{-1} \frac{a\sin \theta}{a\sqrt{1 - \sin^2 \theta}}$

$=\tan^{-1} \left ( \frac{\sin \theta}{\sqrt{\cos^2 \theta}} \right ) = \tan^{-1} \left ( \frac{\sin \theta}{{\cos \theta}} \right )$

$=\tan^{-1}\left ( \tan \theta \right )$

$=\theta = \sin^{-1}\left ( \frac{x}{a} \right )$  is the simplest form.

Given $\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right )$

Here we can take $x = a\tan \theta \Rightarrow \frac{x}{a} = \tan \theta$

So, $\theta = \tan^{-1}\left ( \frac{x}{a} \right )$

$\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right )$ will become;

$=\tan^{-1}\left(\frac{3a^2a\tan \theta -(a\tan \theta)^3}{a^3 - 3a(a\tan \theta)^2} \right ) = \tan^{-1}\left(\frac{3a^3\tan \theta -a^3\tan ^3 \theta}{a^3 - 3a^3\tan ^2 \theta} \right )$

and as $\left [ \because \left(\frac{3\tan \theta -\tan ^3 \theta}{ 1- 3\tan ^2 \theta} \right) =\tan 3\theta \right ]$;

$=3 \theta$

$=3 \tan^{-1}(\frac{x}{a})$

hence the simplest form is $3 \tan^{-1}(\frac{x}{a})$.

Given equation:

$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ]$

So, solving the inner bracket first, we take the value of $\sin x^{-1} \frac{1}{2} = x.$

Then we have,

$\sin x = \frac{1}{2} = \sin \left ( \frac{\pi}{6} \right )$

Therefore, we can write $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$.

$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ] = \tan^{-1}\left[2\cos\left(2\times\frac{\pi}{6} \right ) \right ]$

$= \tan^{-1}\left[2\cos\left(\frac{\pi}{3} \right ) \right ] = \tan^{-1}\left[2\times\left(\frac{1}{2} \right ) \right ] = \tan^{-1}1 = \frac{\pi}{4}$.

We have to find the value of   $\cot(\tan^{-1}a + \cot^{-1}a)$

As we know $\left [\because \tan^{-1}x + \cot^{-1} x = \frac{\pi}{2} \right ]$   so,

Equation reduces to $\cot(\frac{\pi}{2}) = 0$.

Taking the value $x = \tan \Theta$  or  $\tan^{-1}x = \Theta$  and $y = \tan \Theta$  or  $\tan^{-1} y = \Theta$then we have,

$\tan \frac{1}{2}\left[\sin^{-1}\frac{2\tan \Theta}{1+(\tan \Theta)^2} + cos^{-1}\frac{1-\tan^2 \Theta}{1+(\tan \Theta)^2} \right ]$,

$\tan \frac{1}{2}\left[\sin^{-1}(\sin2\Theta) + cos^{-1} (\cos 2\Theta) \right ]$

$\because \left[\cos^{-1}(\frac{1-\tan^2 \Theta}{1+ \tan^2\Theta}) = \cos^{-1} (\cos2 \Theta) , \right ]$

$\because \left[\sin^{-1}(\frac{2\tan\Theta}{1+ \tan^2\Theta}) = \sin^{-1} (\sin2 \Theta) \right ]$

Then,

$=\tan \frac{1}{2}\left[2\tan^{-1}x + 2\tan^{-1}y \right ]$  $\because \left[\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right]$

$=\tan \left [ \tan^{-1}\frac{x+y}{1-xy} \right ]$

$=\frac{x+y}{1-xy}$  Ans.

As we know the identity;

$sin^{-1} x + cos^{-1} x = \frac {\pi}{2},\ x\ \epsilon\ [-1,1]$. it will just hit you by practice to apply this.

So, $\sin\left(\sin^{-1}\frac{1}{5} + \cos ^{-1}x \right ) =1$     or    $\sin^{-1}\frac{1}{5} + \cos ^{-1}x =\sin^{-1}(1)$,

we can then write $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$,

putting in above equation we get;

$\sin^{-1}\frac{1}{5} + \frac{\pi}{2} - \sin^{-1}x =\frac{\pi}{2}$                          $\because \left [ \sin^{-1}(1)=\frac{\pi}{2} \right ]$

$\sin^{-1}x = \sin^{-1} \frac{1}{5}$

Ans.$x = \frac{1}{5}$

Using the identity $\tan^{-1}x+\tan^{-1} y = \tan^{-1}{\frac{x+y}{1-xy}}$,

We can find the value of x;

So, $\tan^{-1}\frac{x-1}{x-2} + \tan^{-1}\frac{x+1}{x+2} =\frac{\pi}{4}$

on applying,

$\tan^{-1}{\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1- \left ( \frac{x-1}{x-2} \right )\left ( \frac{x+1}{x+2} \right )}}$

$=\tan^{-1}\frac{\frac{(x-1)(x+2)+(x-2)(x+1)}{x^2-4}}{1-\frac{x^2-1}{x^2-4}} = \tan^{-1} \left [ \frac{2x^2-4}{-3} \right ] = \frac{\pi}{4}$

$=\frac{2x^2-4}{-3} = \tan (\frac{\pi}{4})=1$

$2x^2=1$    or   $x = \pm \frac{1}{\sqrt{2}}$,

Hence, the possible values of x are $\pm \frac{1}{\sqrt{2}}$.

Given $\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$;

We know that $\sin^{-1}(\sin x) = x$

If the value of x belongs to  $\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$ then we get the principal values of $\sin^{-1}x$.

Here, $\frac{2\pi}{3} \notin \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$

We can write $\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$  is as:

$\sin^{-1}\left [ \sin\left ( \pi-\frac{2\pi}{3} \right ) \right ]$

= $\sin^{-1}\left [ \sin \frac{\pi}{3} \right ]$  where $\frac{\pi}{3} \epsilon \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ]$

$\therefore \sin^{-1}\left (\sin\frac{2\pi}{3} \right )=\sin^{-1}\left [ \sin \frac{\pi}{3} \right ]=\frac{\pi}{3}$

As we know $\tan^{-1}\left ( \tan x \right ) =x$

If $x \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ).$ which is the principal value range of $\tan^{-1}x$.

So, as in $\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$;

$\frac{3\pi}{4}\notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

Hence we can write $\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$   as :

$\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$ = $\tan^{-1}\left (\tan\frac{3\pi}{4} \right) = \tan^{-1}\left [ \tan(\pi - \frac{\pi}{4}) \right ] = \tan^{-1}\left [ \tan (\frac{-\pi}{4}) \right ]$

Where $-\frac{\pi}{4} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

and $\therefore \tan^{-1}\left (\tan\frac{3\pi}{4} \right )=\tan^{-1}\left [ \tan (\frac{-\pi}{4}) \right ]=-\frac{\pi}{4}$

Given that $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$

we can take $\sin^{-1}\frac{3}{5} = x$,

then  $\sin x = \frac{3}{5}$

or  $\cos x = \sqrt{1-\sin^{2}x}= \frac{4}{5}$

$\Rightarrow \tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

$\Rightarrow \tan^{-1}\frac{3}{4}= x$

We have similarly;

$\cot^{-1} \frac{3}{2} = \tan^{-1} \frac{2}{3}$

Therefore we can write $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$

$=\tan\left(\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{2}{3} \right )$

$=\tan\left[\tan^{-1}\left ( \frac{\frac{3}{4}+\frac{2}{3}}{1- \frac{3}{4}.\frac{2}{3}} \right ) \right ]$            from   $As, \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]$

$=\tan \left (\tan^{-1} \frac{9+8}{12-6} \right ) = \tan \left (\tan^{-1} \frac{17}{6} \right )= \frac{17}{6}$

(A)    $\frac{7\pi}{6}$

(B)    $\frac{5\pi}{6}$

(C)    $\frac{\pi}{3}$

(D)    $\frac{\pi}{6}$

As we know that $\cos^{-1} (cos x ) = x$ if $x\epsilon [0,\pi]$ and is principal value range of $\cos^{-1}x$.

In this case $\cos^{-1}\left(\cos\frac{7\pi}{6} \right )$,

$\frac{7\pi}{6} \notin [0,\pi]$

hence we have then,

$\cos^{-1}\left(\cos\frac{7\pi}{6} \right ) =$  $\cos^{-1} \left ( \cos \frac{-7\pi}{6} \right ) = \cos^{-1}\left [ \cos\left ( 2\pi - \frac{7\pi}{6} \right ) \right ]$

$\left [ \because \cos (2\pi + x) = \cos x \right ]$

$\therefore\ we\ have \cos^{-1}\left ( \cos \frac{7\pi}{6} \right ) = \cos^{-1}\left ( \cos \frac{5\pi}{6} \right ) = \frac{5\pi}{6}$

Hence the correct answer is $\frac{5\pi}{6}$ (B).

(A)    $\frac{1}{2}$

(B)

(C)    $\frac{1}{4}$

(D)    $1$

Solving the inner bracket of $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$;

$\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$  or

Take $\sin^{-1}\left(-\frac{1}{2} \right ) = x$  then,

$\sin x =-\frac{1}{2}$  and we know the range of principal value of $\sin^{-1}x\ is\ \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ].$

Therefore we have $\sin^{-1}\left ( -\frac{1}{2} \right ) = -\frac{\pi}{6}$.

Hence, $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right ) = \sin \left ( \frac{\pi}{3}+ \frac{\pi}{6} \right )= \sin \left ( \frac{3\pi}{6} \right ) = \sin\left ( \frac{\pi}{2} \right ) = 1$

Hence the correct answer is D.

(A)    $\pi$

(B)    $-\frac{\pi}{2}$

(C)    0

(D)    $2\sqrt3$

We have $\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)$;

finding the value of  $\cot^{-1}(-\sqrt3)$:

Assume $\cot^{-1}(-\sqrt3) =y$ then,

$\cot y = -\sqrt 3$  and the range of the principal value of $\cot^{-1}$ is $(0,\pi)$.

Hence, principal value is $\frac{5\pi}{6}$

Therefore $\cot^{-1} (-\sqrt3) = \frac {5\pi}{6}$

and $\tan^{-1} \sqrt3 = \frac{\pi}{3}$

so, we have now,

$\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)=\frac{\pi}{3} - \frac{5\pi}{6}$

$= \frac{2\pi - 5\pi}{6} = \frac{-3\pi}{6}$

or, $= \frac{ -\pi}{2}$

Hence the answer is option  (B).

### Solutions of NCERT for class 12 maths chapter 2 Inverse Trigonometric Functions: Miscellaneous Exercise

If $x \epsilon [0,\pi]$  then  $\cos^{-1}(\cos x) = x$ , which is principal value of $\cos^{-1} x$.

So, we have $\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )$

$where \:\frac{13\pi}{6} \notin \left [ 0, \pi \right ].$

$Hence\: we \:can\: write\: \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) \:as$

$=\cos^{-1}\left (\cos\left(2\pi+\frac{\pi}{6} \right ) \right )$

$=\cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right )$

$\frac{\pi}{6}\ \epsilon \left [ 0, \pi \right ]$

Therefore we have,

$\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) = \cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right ) = \frac{\pi}{6}$.

We have given $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$;

so, as we know  $\tan^{-1}\left(\tan x \right ) =x \:\:if\:\:x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

So, here we have $\frac{7\pi}{6} \notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$.

Therefore we can write $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$ as:

$=\tan^{-1}\left(\tan \left (2\pi - \frac{5\pi}{6} \right ) \right )$           $\left [ \because \tan(2\pi - x) = -\tan x \right ]$

$=\tan^{-1}\left[-\tan \left ( \frac{5\pi}{6} \right ) \right ]$

$=\tan^{-1}\left[\tan \left (\pi- \frac{5\pi}{6} \right ) \right ]$

$=\tan^{-1}\left[\tan \left (\frac{\pi}{6} \right ) \right ]\:\:where\:\:\frac{\pi}{6} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

$\therefore \tan^{-1} \left ( \tan \frac{7\pi}{6} \right ) = \tan^{-1} \left ( \tan \frac{\pi}{6} \right ) = \frac{\pi}{6}$.

To prove: $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$;

$L.H.S=2\sin^{-1}\frac{3}{5}$

Assume that  $\sin^{-1}\frac{3}{5} = x$

then we have $\sin x = \frac{3}{5}$.

or $\cos x = \sqrt{1-\left (\frac{3}{5} \right )^2} = \frac{4}{5}$

Therefore we have

$\tan x = \frac{3}{4}\:\:or\:\:x = \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$

Now,

We can write L.H.S as

$2\sin^{-1}\frac{3}{5} = 2\tan^{-1}\frac{3}{4}$

$=\tan^{-1} \left [\frac{2\times\frac{3}{4}}{1- \left ( \frac{3}{4} \right )^2} \right ]$    as we know    $\left [2\tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \right ]$

$=\tan^{-1} \left [\frac{\frac{3}{2}}{\left ( \frac{16-9}{16} \right )} \right ] = \tan^{-1}\left ( \frac{3}{2}\times \frac{16}{7} \right )$

$=\tan^{-1} \frac{24}{7}=R.H.S$

L.H.S = R.H.S

Taking $\sin ^{-1} \frac{8}{17} = x$

then,

$\sin x = \frac{8}{17} \Rightarrow \cos x = \sqrt{1- \left ( \frac{8}{17} \right )^2} = \sqrt {\frac{225}{289}} = \frac{15}{17}.$

Therefore we have-

$\tan^{-1} x = \frac{8}{15} \Rightarrow x = \tan^{-1} \frac{8}{15}$

$\therefore \sin ^{-1} \frac{8}{17} = \tan ^{-1} \frac{8}{15}$        .............(1).

$Now, let\:\sin ^{-1} \frac{3}{5} = y$,

Then,

$\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{3}{4}$                    .............(2).

So, we have now,

L.H.S.

$\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5}$

using equations (1) and (2) we get,

$=\tan ^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}$

$=\tan^{-1} \frac{\frac{8}{15}+ \frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}$                   $[\because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1 - xy}]$

$=\tan^{-1} (\frac{32+45}{60-24})$

$=\tan^{-1} (\frac{77}{36})$

=  R.H.S.

Take $\cos^{-1}\frac{4}{5} = x$  and $\cos^{-1}\frac{12}{13} = y$    and $\cos^{-1}\frac{33}{65} = z$

then we have,

$\cos x = \frac{4}{5}$

$\sin x = \sqrt {1- \left ( \frac {4}{5} \right )^2} = \frac {3}{5}$

Then we can write it as:

$\tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$    or    $x= \tan^{-1} \frac{3}{4}$

$\therefore \cos ^{-1} \frac{4}{5} = \tan^{-1} \frac{3}{4}$                                   ...............(1)

Now, $\cos^{-1}\frac{12}{13} = y$

$\cos y = \frac{12}{13} \Rightarrow$ $\sin y =\frac{5}{13}$

$\therefore \tan y = \frac{5}{12} \Rightarrow y = \tan^{-1} \frac{5}{12}$

So,  $\cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}$                        ...................(2)

Also we have similarly;

$\cos^{-1}\frac{33}{65} = z$

Then,

$\cos^{-1}\frac{33}{65} = \tan^{-1} \frac{56}{33}$             ...........................(3)

Now, we have

L.H.S

$\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13}$    so, using (1) and (2) we get,

$=\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{5}{12}$

$=\tan^{-1}\left ( \frac{\frac{3}{4}+ \frac{5}{12}}{1-\left ( \frac{3}{4}\times \frac{5}{12} \right )} \right )$             $\because \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]$

$=\tan^{-1}\left ( \frac{36+20}{48-15} \right )$

$=\tan^{-1}\left ( \frac{56}{33} \right )$    or we can write it as;

$=\cos^{-1}\frac{33}{65}$

=  R.H.S.

Hence proved.

Converting all terms in tan form;

Let $\cos^{-1}\frac{12}{13} = x$ ,  $\sin^{-1}\frac{3}{5} = y$  and $\sin^{-1}\frac{56}{65} = z$.

now, converting all the terms:

$\cos^{-1}\frac{12}{13} = x$  or   $\cos x = \frac{12}{13}$

We can write it in tan form as:

$\cos x = \frac{12}{13} \Rightarrow$ $\sin x = \frac{5}{13}$.

$\therefore \tan x = \frac{5}{12} \Rightarrow x = \tan^{-1} \frac{5}{12}$

or $\cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}$           ................(1)

$\sin^{-1}\frac{3}{5} = y$     or   $\sin y = \frac{3}{5}$

We can write it in tan form as:

$\sin y = \frac{3}{5} \Rightarrow$$\cos y = \frac{4}{5}$

$\therefore \tan y =\frac{3}{4} \Rightarrow y = \tan^{-1} \frac{3}{4}$

or $\sin^{-1}\frac{3}{5} = \tan^{-1} \frac{3}{4}$         ......................(2)

Similarly, for $\sin^{-1}\frac{56}{65} = z$;

we have $\sin^{-1}\frac{56}{65} = \tan^{-1} \frac{56}{33}$     .............(3)

Using (1) and (2) we have L.H.S

$\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5}$

$= \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{3}{4}$

On applying $\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}$

We have,

$=\tan^{-1} \frac{\frac{5}{12}+\frac{3}{4}}{1-(\frac{5}{12}.\frac{3}{4})}$

$=\tan^{-1} (\frac{20+36}{48-15})$

$=\tan^{-1} (\frac{56}{33})$

$=\sin^{-1} (\frac{56}{65})$                               ...........[Using (3)]

=R.H.S.

Hence proved.

Taking R.H.S;

We have $\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}$

Converting sin and cos terms in tan forms:

Let $\sin^{-1}\frac{5}{13} = x$   and   $\cos^{-1}\frac{3}{5} = y$

now, we have $\sin^{-1}\frac{5}{13} = x$   or    $\sin x = \frac{5}{13}$

$\sin x = \frac{5}{13} \:or\: \cos x =\frac{12}{13}\:or\:\tan x = \frac{5}{12}$

$\tan x = \frac{5}{12} \Rightarrow x =\tan^{-1} \frac{5}{12}$

$\therefore \sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12}$                 ............(1)

Now, $\cos^{-1}\frac{3}{5} = y\Rightarrow \cos y = \frac{3}{5}$

$\cos y = \frac{3}{5} \:or\: \sin y = \frac{4}{5}\:or\:\tan y = \frac{4}{3}$

$\tan y = \frac{4}{3} \Rightarrow y = \tan^{-1} \frac{4}{3}$

$\therefore \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{4}{5}$                 ................(2)

Now, Using (1) and (2) we get,

R.H.S.

$\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}$

$=\tan^{-1}\left ( \frac{\frac{5}{12}+\frac{4}{3}}{1- \frac{5}{12}\times \frac{4}{3}} \right )$    as we know $\left [ \tan^{-1} x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy} \right ]$

so,

$= \tan^{-1} \frac{63}{16}$

equal to L.H.S

Hence proved.

Applying the formlua:

$\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}$    on two parts.

we will have,

$=\tan^{-1}\left (\frac{\frac{1}{5}+ \frac{1}{7}}{1- \frac{1}{5}\times \frac{1}{7}} \right ) + \tan^{-1}\left (\frac{\frac{1}{3}+ \frac{1}{8}}{1- \frac{1}{3}\times \frac{1}{8}} \right )$

$= \tan^{-1} \left ( \frac{7+5}{35-1} \right ) + \tan^{-1} \left ( \frac{8+3}{24-1} \right )$

$= \tan^{-1} \left ( \frac{12}{34} \right ) + \tan^{-1} \left ( \frac{11}{23} \right )$

$= \tan^{-1} \left ( \frac{6}{17} \right ) + \tan^{-1} \left ( \frac{11}{23} \right )$

$= \tan^{-1}\left [ \frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}\times\frac{11}{23}} \right ]$

$= \tan^{-1}\left [ \frac{325}{325} \right ] = \tan^{-1} 1$

$=\frac{\pi}{4}$

Hence it s equal to R.H.S

Proved.

By observing the square root we will first put

$x= \tan^2 \theta$.

Then,

we have $\tan^{-1} \sqrt{\tan^2 \theta} = \frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta}$

or, R.H.S.

$\frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \frac{1}{2}\cos^{-1}(cos2 \theta)$

$= \frac{1}{2}\times 2\theta = \theta$.

L.H.S. $\tan^{-1} \sqrt{\tan^2 \theta} = \tan^{-1}(\tan \theta) = \theta$

hence L.H.S. = R.H.S proved.

Given that $\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )$

By observing we can rationalize the fraction

$\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )$

We get then,

$=\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right ) = \left(\frac{(\sqrt{1+\sin x} + \sqrt{1 - \sin x})^2}{{1+\sin x} - {1 + \sin x}} \right )$

$= \left(\frac{{1+\sin x} +{1 - \sin x} + 2\sqrt{(1+\sin x)(1-\sin x)} }{{1+\sin x} - {1 + \sin x}} \right )$

$= \frac{2(1+\sqrt{1-\sin^2 x})}{2\sin x} = \frac{1+\cos x}{\sin x} = \frac{2\cos^2 \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}$

$= \cot \frac{x}{2}$

Therefore we can write it as;

$\cot^{-1}\left ( \cot \frac{x}{2} \right ) = \frac{x}{2}$

As L.H.S. = R.H.S.

Hence proved.

[Hint: Put $x = \cos 2\theta$]

By using the Hint we will put $x = \cos 2\theta$;

we get then,

$=\tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right )$

$=\tan^{-1}\left(\frac{\sqrt{2\cos^2 \theta} - \sqrt{2\sin^2\theta}}{\sqrt{2\cos^2 \theta} + \sqrt{2\sin^2\theta}} \right )$

$=\tan^{-1}\left(\frac{\sqrt2{\cos \theta} - \sqrt2{\sin\theta}}{\sqrt2{\cos \theta} + \sqrt2{\sin\theta}} \right )$

$=\tan^{-1}\left(\frac{{\cos \theta} - {\sin\theta}}{{\cos \theta} + {\sin\theta}} \right )$   dividing numerator and denominator by $\cos \theta$,

we get,

$= \tan^{-1}\left ( \frac{1-\tan \theta}{1+\tan \theta} \right )$

$= \tan^{-1} 1 - \tan^{-1} (\tan \theta)$      using the formula  $\left [ \tan^{-1}x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy} \right ]$

$= \frac{\pi}{4} - \theta = \frac{\pi}{4}- \frac{1}{2}\cos^{-1}x$

As L.H.S = R.H.S

Hence proved

We have to solve the given equation:

$\frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}$

Take $\frac{9}{4}$ as common in L.H.S,

$=\frac{9}{4}\left [ \frac{\pi}{2}- \sin^{-1}\frac{1}{3} \right ]$

or $=\frac{9}{4}\left [ \cos^{-1}\frac{1}{3} \right ]$    from      $\left [ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \right ]$

Now, assume,

$\left [ \cos^{-1}\frac{1}{3} \right ] = y$

Then,

$\cos y = \frac{1}{3} \Rightarrow \sin y = \sqrt{1-(\frac{1}{3})^2} = \frac{2.\sqrt2}{3}$

Therefore we have now,

$y = \sin^{-1} \frac{2.\sqrt2}{3}$

So we have L.H.S then $= \frac{9}{4}\sin^{-1} \frac{2.\sqrt2}{3}$

That is equal to R.H.S.

Hence proved.

Given equation $2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)$;

Using the formula:

$\left [ 2\tan^{-1}z = \tan^{-1} \frac{2z}{1-z^2} \right ]$

We can write

$2\tan^{-1}(\cos x) = \tan^{-1}\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ]$

$\tan^{-1}\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \tan^{-1}\left [2cosec x \right ]$

So, we can equate;

$=\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \left [2cosec x \right ]$

$=\left [ \frac{2\cos x}{\sin^2 x }\right ] = \left [ \frac{2}{sinx } \right ]$

that implies that $\cos x = \sin x$.

or    $\tan x =1$      or    $x = \frac{\pi}{4}$

Hence we have solution $x = \frac{\pi}{4}$.

Given equation is

$\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x$:

L.H.S can be written as;

$\tan^{-1} \frac{1-x}{1+x} = \tan^{-1}1 - \tan^{-1}x$

Using the formula $\left [ \tan^{-1}x -\tan^{-1}y = \tan^{-1} \frac{x-y}{1+xy} \right ]$

So, we have $\tan^{-1}1 - \tan^{-1}x = \frac{1}{2} \tan^{-1}x$

$\Rightarrow \tan^{-1}1= \frac{3}{2} \tan^{-1}x$

$\Rightarrow\frac{\pi}{4}= \frac{3}{2} \tan^{-1}x$

$\Rightarrow \tan^{-1}x = \frac{\pi}{6}$

$\Rightarrow x= \tan \frac{\pi}{6} = \frac{1}{\sqrt3}$

Hence the value of $x= \frac{1}{\sqrt3}$.

(A)    $\frac{x}{\sqrt{1-x^2}}$

(B)    $\frac{1}{\sqrt{1-x^2}}$

(C)    $\frac{1}{\sqrt{1+x^2}}$

(D)    $\frac{x}{\sqrt{1+x^2}}$

Let $\tan^{-1}x = y$ then we have;

$\tan y = x$   or

$y=\sin^{-1} \left ( \frac{x}{\sqrt{1+x^2}} \right ) \Rightarrow \tan^{-1} x = \sin^{-1} \left ( \frac{x}{\sqrt{1+x^2}} \right)$

$\Rightarrow \sin \left ( \tan^{-1} x \right ) = \sin\left ( \sin^{-1}\left ( \frac{x}{\sqrt{1+x^2}} \right ) \right ) = \frac{x}{\sqrt {1+x^2}}$

Hence the correct answer is D.

(A)    $0,\frac{1}{2}$

(B)    $1,\frac{1}{2}$

(C)    0

(D)    $\frac{1}{2}$

Given the equation: $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$

we can migrate the $\sin^{-1}(1-x)$ term to the R.H.S.

then we have;

$- 2\sin^{-1}x = \frac{\pi}{2} - \sin^{-1}(1-x)$

or $- 2\sin^{-1}x =\cos^{-1}(1-x)$                               ............................(1)

from    $\left [\because \cos^{-1}(1-x) + \sin^{-1}(1-x) = \frac{\pi}{2} \right ]$

Take $\sin^{-1}x = \Theta$  $\Rightarrow \sin \Theta = x$    or    $\cos \Theta = \sqrt{1-x^2}$.

So, we conclude that;

$\sin^{-1}x = \cos^{-1}\left ( \sqrt{1-x^2} \right )$

Therefore we can put the value of $\sin^{-1}x$ in equation (1)  we get,

$- 2\cos^{-1}\left ( \sqrt{1-x^2} \right ) =\cos^{-1}(1-x)$

Putting x= sin y, in the above equation; we have then,

$\Rightarrow - 2\cos^{-1}\left ( \sqrt{1-(\sin y)^2} \right ) =\cos^{-1}(1-\sin y )$

$\Rightarrow - 2\cos^{-1}\left ( \sqrt{\cos^2 y} \right ) =\cos^{-1}(1-\sin y )$

$\Rightarrow - 2\cos^{-1}\left ( \cos y \right ) =\cos^{-1}(1-\sin y )$

$\Rightarrow \cos(-2y) = 1-\sin y$

$\Rightarrow - 2y=\cos^{-1}(1-\sin y )$

$\Rightarrow 1- 2\sin^2 y = 1-\sin y$

$\Rightarrow 2\sin^2 y - \sin y = 0$

$\Rightarrow \sin y(2 \sin y -1) = 0$

So, we have the solution;

$\sin y = 0\ or\ \frac{1}{2}$    Therefore we have $x = 0\ or\ x= \frac{1}{2}$.

When we have $x= \frac{1}{2}$, we can see that :

$L.H.S. = \sin ^{-1}\left ( 1 - \frac{1}{2} \right ) - 2\sin^{-1}\frac{1}{2} = - \sin^{-1}\frac{1}{2} = -\frac{\pi}{6}$

So, it is not equal to the R.H.S. $-\frac{\pi}{6} \neq \frac{\pi}{2}$

Thus we have only one solution which is x = 0

Hence the correct answer is (C).

(A)    $\frac{\pi}{2}$

(B)    $\frac{\pi}{3}$

(C)    $\frac{\pi}{4}$

(D)    $\frac{3\pi}{4}$

Applying formula: $\left [ \tan^{-1} x - \tan^{-1}y = \tan^{-1} \left ( \frac{x-y}{1+xy} \right ) \right ]$.

We get,

$\tan^{-1}\left (\frac{x}{y} \right )-\tan^{-1} \left ( \frac{x-y}{x+y} \right ) = \tan^{-1} \left [\frac{ \frac{x}{y} - \frac{x-y}{x+y}}{1+\left ( \frac{x}{y} \right ) \left ( \frac{x-y}{x+y} \right ) } \right ]$

$= \tan^{-1} \left [\frac{ \frac{x}{y} - \frac{x-y}{x+y}}{1+\left ( \frac{x}{y} \right ) \left ( \frac{x-y}{x+y} \right ) } \right ] = \tan^{-1} \left [ \frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y) + x(x-y)}{y(x+y)}} \right ]$

$= \tan^{-1}\left ( \frac{x^2+xy - xy + y^2}{xy + y^2 + x^2 - xy} \right )$

$= \tan^{-1}\left ( \frac{x^2 + y^2}{ y^2 + x^2 } \right ) = \tan^{-1} 1 = \frac{\pi}{4}$

Hence, the correct answer is C.

## NCERT solutions for class 12 maths chapter-wise

 chapter 1 Solutions of NCERT for class 12 maths chapter 1 Relations and Functions chapter 2 NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions chapter 3 NCERT solutions for class 12 maths chapter 3 Matrices chapter 4 CBSE NCERT solutions for class 12 maths chapter 4 Determinants chapter 5 Solutions of NCERT for class 12 maths chapter 5 Continuity and Differentiability chapter 6 CBSE NCERT solutions for class 12 maths chapter 6 Application of Derivatives chapter 7 NCERT solutions for class 12 maths chapter 7 Integrals chapter 8 Solutions of NCERT for class 12 maths chapter 8 Application of Integrals chapter 9 CBSE NCERT solutions for class 12 maths chapter 9 Differential Equations chapter 10 NCERT solutions for class 12 maths chapter 10 Vector Algebra chapter 11 Solutions of NCERT for class 12 maths chapter 11 Three Dimensional Geometry chapter 12 CBSE NCERT solutions for class 12 maths chapter 12 Linear Programming chapter 13 NCERT solutions for class 12 maths chapter 13 Probability

## NCERT solutions for class 12

Students who wish to perform well in the CBSE 12 board examination, solutions of NCERT for class 12 maths chapter 2 inverse trigonometric functions are very helpful but here are some tips to make command on Inverse Trigonometric Functions.

• The inverse trigonometric function is inverse of the trigonometric function, so if you have a command on the trigonometric function then it will be easy for you to understand inverse trigonometric functions.
• Try to relate trigonometric functions formulas with inverse trigonometric functions formulas, so that memorizing the formulae becomes easier.
• Before starting to solve exercise, first solve the examples that are given in the NCERT class 12 maths textbook.
• Also, try to solve every exercise including miscellaneous exercise, NCERT chapter examples, miscellaneous examples on your own, if you are finding difficulties, you can take the help of CBSE NCERT solutions for class 12 maths chapter 2 inverse trigonometric functions.
• If you have solved all NCERT then you can solve previous years paper CBSE board to get familiar with the pattern of board exam question paper

Happy learning!!!