Class 12 Maths Ncert Solutions Pdf Chapter 2

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions - Students appearing in Class 12 board exams must go through these NCERT solutions of Class 12 Maths. These NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions is one of the most important chapters in the NCERT Class 12 Maths book. These class 12 Maths chapter 2 notes will be beneficial for board exams as well as for other entrance exams such as JEE.

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

All these questions are prepared and explained in a step-by-step method in the NCERT solutions for class 12 Maths chapter 2. Read further to know more about the NCERT solutions.

NCERT Solutions for Class 12 Maths Chapter 2 Introduction

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 Class 12 Maths chapter 2 notes.

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

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

Question:1 Find the principal values of the following : $\sin^{-1}\left ( \frac{-1}{2} \right )$

Answer:

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 $sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$

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

Question:2 Find the principal values of the following: $\cos^{-1}\left(\frac{\sqrt3}{2} \right )$

Answer:

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 $\frac{\pi}{6}$ .

Question:5 Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{2} \right )$

Answer:

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.

Question:6 Find the principal values of the following : $\tan^{-1}(-1)$

Answer:

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

$\tan^{-1}(-1) =z$ ,

$\tan z = -1$

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

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 $-\frac{\pi}{4}$ .

Question:8 Find the principal values of the following: $\cot^{-1}(\sqrt3)$

Answer:

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 $\frac{\pi}{6}$

Question:11 Find the values of the following: $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2} \right ) + \sin^{-1}\left(-\frac{1}{2} \right )$

Answer:

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:

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

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

Question:12 Find the values of the following: $\cos^{-1}\left(\frac{1}{2} \right ) + 2\sin^{-1}\left(\frac{1}{2} \right )$

Answer:

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 = $\frac{\pi}{3}+2(\frac{\pi}{6}) = \frac{2\pi}{3}$ .

Question:13 If $\sin^{-1}x = y$ then

(A) $0\leq y \leq \pi$

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

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

Answer:

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.

Question:14 $\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$ is equal to

(A) $\pi$

(B) $-\frac{\pi}{3}$

(D) $\frac{2\pi}{3}$

Answer:

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;

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

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

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

Question:1 Prove the following: $3\sin^{-1}x = \sin^{-1}(3x - 4x^3),\;\;x\in\left[-\frac{1}{2},\frac{1}{2} \right ]$

Answer:

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

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

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

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

= $3\theta$

= $3\sin^{-1}x$ = L.H.S

Question:2 Prove the following: $3\cos^{-1} x = \cos^{-1}(4x^3 - 3x), \;\;x\in\left[\frac{1}{2},1 \right ]$

Answer:

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.

Question:4 Prove the following: $2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}$

Answer:

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.

Question:14 If $\sin\left(\sin^{-1}\frac{1}{5} + \cos ^{-1}x \right ) =1$ , then find the value of $x$ .

Answer:

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

Question:15 If $\tan^{-1}\frac{x-1}{x-2} + \tan^{-1}\frac{x+1}{x+2} =\frac{\pi}{4}$ , then find the value of $x$ .

Answer:

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 $\pm \frac{1}{\sqrt{2}}$ .

Question:19 $\cos^{-1}\left(\cos\frac{7\pi}{6} \right )$ is equal to

(A) $\frac{7\pi}{6}$

(B) $\frac{5\pi}{6}$

(D) $\frac{\pi}{6}$

Answer:

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).

Question:20 $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$ is equal to

(A) $\frac{1}{2}$

(B)

(D) $1$

Answer:

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.

Question:21 $\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)$ is equal to

(A) $\pi$

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

(D) $2\sqrt3$

Answer:

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).

NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions: Miscellaneous Exercise

Question:1 Find the value of the following: $\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )$

Answer:

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,

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

Question:3 Prove that $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$

Answer:

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

Question:4 Prove that $\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} =\tan^{-1}\frac{77}{36}$

Answer

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.

Question:5 Prove that $\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}$

Answer:

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.

Question:6 Prove that $\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}$

Answer:

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.

Question:7 Prove that $\tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}$

Answer:

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.

Question:8 Prove that $\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} +\tan^{-1}\frac{1}{3} +\tan^{-1}\frac{1}{8} = \frac{\pi}{4}$

Answer:

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.

Question:9 Prove that $\tan^{-1} \sqrt{x} = \frac{1}{2}\cos^{-1}\frac{1-x}{1+x},\;\;x\in [0,1]$

Answer:

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.

Question:10 Prove that $\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right ) = \frac{x}{2},\;\;x\in\left(0,\frac{\pi}{4} \right )$

Answer:

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.

Question:11 Prove that $\tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right ) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x,\;\;-\frac{1}{\sqrt2}\leq x\leq 1$

[Hint: Put $x = \cos 2\theta$ ]

Answer:

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

Question:12 Prove that $\frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}$

Answer:

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.

Question:13 Solve the following equations: $2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)$

Answer:

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

Question:16 $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$ then $x$ is equal to

(A) $0,\frac{1}{2}$

(B) $1,\frac{1}{2}$

(D) $\frac{1}{2}$

Answer:

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).

NCERT solutions for class 12 maths - Chapter wise

Tips to use NCERT Solutions for Class 12 Maths Chapter 2

If you wish to perform well in the CBSE 12 board examination, NCERT Class 12 Maths solutions 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, use NCERT solutions for class 12 maths chapter 2 pdf to learn easily.
Try to relate trigonometric functions formulas with inverse trigonometric functions formulas, so that memorizing the formulae becomes easier.
Before starting to solve an exercise, first solve the examples that are given in the NCERT class 12 maths textbook.
Also, try to solve every exercise including inverse trigonometric functions class 12 miscellaneous, and examples, miscellaneous examples on your own, if you are finding difficulties, you can take the help of Class 12 Maths Chapter 2 NCERT solutions.
If you have solved all NCERT then you can solve previous years paper CBSE board to get familiar with the pattern of the board exam question paper.

Also, check NCERT solutions for class 12 for other chapter and subjects.

Class 12 Maths Ncert Solutions Pdf Chapter 2

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Class 12 Maths Ncert Solutions Pdf Chapter 2

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

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

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

NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions: Miscellaneous Exercise

NCERT solutions for class 12 maths - Chapter wise

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