NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Komal MiglaniUpdated on 22 Aug 2025, 03:28 PM IST

In the universe of trigonometry, inverse functions are the key to unlocking the angles when only the ratios are given. Inverse Trigonometric Functions Class 12 solutions deal with functions that help determine the angles of a right triangle when only the ratio of one of the two pairs of sides is given. NCERT solutions for class 12 Maths highlight that for every ratio, there exists a unique angle, just as every answer has a corresponding counter-question; similarly, every sine, cosine, or tangent has an inverse.

Inverse trigonometric functions class 12 NCERT solutions mainly focus on the restrictions on domains and ranges of trigonometric functions that ensure the existence of their inverses. This chapter's learning applies to many fields, including engineering, navigation, astronomy, architecture, and robotics. Experienced Careers360 experts prepared these solutions using the NCERT, following the latest CBSE guidelines.

This Story also Contains

  1. NCERT Solution for Class 12 Maths Chapter 2 Solutions: Download PDF
  2. NCERT Solutions for Class 12 Maths Chapter 2: Exercise Questions
  3. Class 12 Maths NCERT Chapter 2: Extra Question
  4. Inverse Trigonometric Functions Class 12 Chapter 2: Topics
  5. Inverse Trigonometric Functions Class 12 NCERT Solutions - Important Formulae
  6. Approach to Solve Questions of Inverse Trigonometric Functions Class 12
  7. What Extra Should Students Study Beyond the NCERT for JEE?
  8. NCERT Solutions for Class 12 Maths: Chapter Wise
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

NCERT Solution for Class 12 Maths Chapter 2 Solutions: Download PDF

Students who wish to access the NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions can click on the link below to download the complete solution in PDF.

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NCERT Solutions for Class 12 Maths Chapter 2: Exercise Questions

NCERT Inverse Trigonometric Functions Class 12 Solutions: Exercise 2.1
Page number: 26-27
Total questions: 14

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 the inverse of 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 3: Find the principal values of the following: $\textup{cosec}^{-1}(2)$

Answer:

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

Question 4: Find the principal values of the following: $\tan^{-1}(-\sqrt3)$

Answer:

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

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 $\tan^{-1}(-1)$, 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 $-\frac{\pi}{4}$ .

Question 7: Find the principal values of the following: $\sec^{-1}\left (\frac{2}{\sqrt3}\right)$

Answer:

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

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 9: Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{\sqrt2} \right )$

Answer:

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 $\frac{3\pi}{4}$ .

Question 10: Find the principal values of the following: $\textup{cosec}^{-1}(-\sqrt2)$

Answer:

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

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 +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) $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

(C) $0 < y < \pi$

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

(C) $\frac{\pi}{3}$

(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 Inverse Trigonometric Functions Class 12 Solutions: Exercise 2.2
Page number: 29-30
Total questions: 15

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 3: Write the following functions in the simplest form: $\tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x},\;\;x\neq 0$

Answer:

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.

Question 4: Write the following functions in the simplest form: $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi$

Answer:

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 $\frac{x}{2}$

Question 5: Write the following functions in the simplest form: $\tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right ),\;\; \frac{-\pi}{4} < x < \frac{3\pi}{4}$

Answer:

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 the numerator and the 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.

Question 6: Write the following functions in the simplest form: $\tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}, \;\; |x| < a$

Answer:

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.

Question 7: Write the following functions in the simplest form: $\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right ),\;\;a>0\;\;;\;\;\frac{-a}{\sqrt3} < x < \frac{a}{\sqrt3}$

Answer:

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

Question 8: Find the values of each of the following: $\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ]$

Answer:

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

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

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

Question 9: Find the values of each of the following: $\tan \frac{1}{2}\left[\sin^{-1}\frac{2x}{1+x^2} + cos^{-1}\frac{1-y^2}{1+y^2} \right ],\;\;|x|<1,\;y>0$ and $xy<1$

Answer:

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,

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

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

$=\frac{x+y}{1-xy}$

Question 10: 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 11: Find the values of each of the expressions$\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$

Answer:

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

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

Question 12: Find the values of each of the expressions $\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$

Answer:

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

Question 13: Find the values of each of the expressions $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$

Answer:

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 similarities

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

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

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

Question 14: $\cos^{-1}\left(\cos\frac{7\pi}{6} \right )$ is equal to

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

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

(C) $\frac{\pi}{3}$

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

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

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

(C) $\frac{1}{4}$

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

(A) $\pi$

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

(C) 0

(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 Inverse Trigonometric Functions Class 12 Solutions: Miscellaneous Exercise
Page number: 31-32
Total questions: 14

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 2: Find the value of the following: $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$

Answer:

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:

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

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

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

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

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

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 similarities;

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

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 to 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 $\tan^{-1} \sqrt{x} = \frac{1}{2}\cos^{-1}\frac{1-x}{1+x},\;\;x\in [0,1]$

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 9: 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 rationalise the fraction

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

We get,

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

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

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

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

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 10: 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,

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

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

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

$=\tan^{-1}\left(\frac{{\cos \theta} - {\sin\theta}}{{\cos \theta} + {\sin\theta}} \right )$ 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 11: 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 12: Solve the following equations: $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x,\;(x>0)$

Answer:

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 $x= \frac{1}{\sqrt3}$ .

Question 13: $\sin(\tan^{-1}x),\;|x|<1$ is equal to

(A) $\frac{x}{\sqrt{1-x^2}}$

(B) $\frac{1}{\sqrt{1-x^2}}$

(C) $\frac{1}{\sqrt{1+x^2}}$

(D) $\frac{x}{\sqrt{1+x^2}}$

Answer:

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.

Question 14: $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$ then $x$ is equal to

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

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

(C) 0

(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).

Also read,

Inverse Trigonometric Functions Class 12 Exercise 2.1

Inverse Trigonometric Functions Class 12 Exercise 2.2

Inverse Trigonometric Functions Class 12 Miscellaneous Exercise

Class 12 Maths NCERT Chapter 2: Extra Question

Question: If $3 \tan ^{-1} x+\cot ^{-1} x=\pi$, then x equals to:

Solution:
Given That, $3 \tan ^{-1} x+\cot ^{-1} x=\pi$
$
\begin{aligned}
& \Rightarrow 2 \tan ^{-1} x+\tan ^{-1} x+\cot ^{-1} x=\pi \\
& \Rightarrow 2 \tan ^{-1} x=\pi-\frac{\pi}{2}\left[\text { Since, } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right] \\
& \Rightarrow \tan ^{-1} \frac{2 x}{1-x^2}=\frac{\pi}{2}\left[\text { Since, } 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^2}\right]
\end{aligned}
$
$
\Rightarrow \frac{2 x}{1-x^2}=\tan \frac{\pi}{2}
$
$
\Rightarrow \frac{2 x}{1-x^2}=\tan \frac{1}{0}
$
Cross multiplying

$
\begin{aligned}
& \Rightarrow 1-x^2=0 \\
& \Rightarrow x^2= \pm 1
\end{aligned}
$
Here, only $\mathrm{x}=1$ satisfies the given equation.
Note: By putting $x=-1$ in the given equation, we get:

$
\begin{aligned}
& 3 \tan ^{-1}(-1)+\cot ^{-1}(-1)=\pi \\
& ⇒3 \tan ^{-1}\left[\tan \left(\frac{-\pi}{4}\right)\right]+\cot ^{-1}\left[\cot \left(\frac{-\pi}{4}\right)\right]=\pi \\
&⇒ 3 \tan ^{-1}\left[-\tan \left(\frac{\pi}{4}\right)\right]+\cot ^{-1}\left[-\cot \left(\frac{\pi}{4}\right)\right]=\pi \\
& ⇒3 \tan ^{-1}\left[\tan \left(\frac{\pi}{4}\right)\right]+\pi-\cot ^{-1}\left[\cot \left(\frac{\pi}{4}\right)\right]=\pi \\
&⇒ -3 \times \frac{\pi}{4}+\pi-\frac{\pi}{4}=\pi \\
&⇒ -\pi+\pi=\pi \\
&⇒ 0 \neq \pi
\end{aligned}
$
$\therefore$ x = -1 does not satisfy the given equation.

Hence, the correct answer is 1.

Inverse Trigonometric Functions Class 12 Chapter 2: Topics

Here is the list of important topics that are covered in Class 12 Chapter 2, Inverse Trigonometric Functions:

  • 2.1 Introduction
  • 2.2 Basic Concepts
  • 2.3 Properties of Inverse Trigonometric Functions

Inverse Trigonometric Functions Class 12 NCERT Solutions - Important Formulae

The inverse of the sine function: sin-1(x) or arcsin(x) is defined on [-1, 1].

Properties of Inverse Trigonometric Functions:

Function

Domain

Range

y = sin-1(x)

[-1, 1]

[-π/2, π/2]

y = cos-1(x)

[-1, 1]

[0, π]

y = cosec-1(x)

R - (-1, 1)

[-π/2, π/2] - {0}

y = sec-1(x)

R - (-1, 1)

[0, π] - {π/2}

y = tan-1(x)

R

(-π/2, π/2)

y = cot-1(x)

R

(0, π)

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Self-Adjusting Trigonometric Properties

  1. $\sin \left(\sin ^{-1}(x)\right)=x$ and $\sin ^{-1}(\sin (x))=x$
  2. $\cos \left(\cos ^{-1}(\mathrm{x})\right)=\mathrm{x}$ and $\cos ^{-1}(\cos (\mathrm{x}))=\mathrm{x}$
  3. $\tan \left(\tan ^{-1}(\mathrm{x})\right)=\mathrm{x}$ and $\tan ^{-1}(\tan (\mathrm{x}))=\mathrm{x}$
  4. $\sec \left(\sec ^{-1}(\mathrm{x})\right)=\mathrm{x}$ and $\sec ^{-1}(\sec (\mathrm{x}))=\mathrm{x}$
  5. $\operatorname{cosec}^{-1}(\operatorname{cosec}(\mathrm{x}))=\mathrm{x}$ and $\operatorname{cosec}\left(\operatorname{cosec}^{-1}(\mathrm{x})\right)=\mathrm{x}$
  6. $\cot ^{-1}(\cot (\mathrm{x}))=\mathrm{x}$ and $\cot \left(\cot ^{-1}(\mathrm{x})\right)=\mathrm{x}$

Reciprocal Relations

  1. $\sin ^{-1}(\frac 1x)=\operatorname{cosec}^{-1}(x)$, where $x \geq 1$ or $x \leq-1$.
  2. $\cos ^{-1}(\frac 1x)=\sec ^{-1}(x)$, Valid for $x \geq 1$ or $x \leq-1$.
  3. $\tan ^{-1}(1\frac 1x)=\cot ^{-1}(x)$, Applies for $x>0$.

Even and Odd Functions

1. $\sin ^{-1}(-x)=-\sin ^{-1}(x)$, for $x \in[-1,1]$
2. $\tan ^{-1}(-x)=-\tan ^{-1}(x)$, for $x \in \mathbb{R}$
3. $\operatorname{cosec} ^{-1}(-x)=-\operatorname{cosec} ^{-1}(x)$, for $|x| \geq 1$
4. $\cos ^{-1}(-x)=\pi-\cos ^{-1}(x)$, for $x \in[-1,1]$
5. $\sec ^{-1}(-x)=\pi-\sec ^{-1}(x)$, for $|x| \geq 1$
6. $\cot ^{-1}(-x)=\pi-\cot ^{-1}(x)$, for $x \in \mathbb{R}$

Complementary Relations

1. $\sin ^{-1}(x)+\cos ^{-1}(x)=\frac{\pi}{2}$
2. $\tan ^{-1}(x)+\cot ^{-1}(x)=\frac{\pi}{2}$
3. $\operatorname{cosec} ^{-1}(x)+\sec ^{-1}(x)=\frac{\pi}{2}$

Sum and Difference Formulas

1. $\tan ^{-1}(x)+\tan ^{-1}(y)=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
2. $\tan ^{-1}(x)-\tan ^{-1}(y)=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$
3. $\sin ^{-1}(x)+\sin ^{-1}(y)=\sin ^{-1}\left[x \sqrt{1-y^2}+y \sqrt{1-x^2}\right]$
4. $\sin ^{-1}(x)-\sin ^{-1}(y)=\sin ^{-1}\left[x \sqrt{1-y^2}-y \sqrt{1-x^2}\right]$
5. $\cos ^{-1}(x)+\cos ^{-1}(y)=\cos ^{-1}\left[x y-\sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right]$
6. $\cos ^{-1}(x)-\cos ^{-1}(y)=\cos ^{-1}\left[x y+\sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right]$
7. $\cot ^{-1}(x)+\cot ^{-1}(y)=\cot ^{-1}\left(\frac{x y-1}{x+y}\right)$
8. $\cot ^{-1}(x)-\cot ^{-1}(y)=\cot ^{-1}\left(\frac{x y+1}{y-x}\right)$

Double Angle Formula

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

Conversion Properties

1. $\sin ^{-1}(x)=\cos ^{-1}\left(\sqrt{1-x^2}\right)=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)=\cot ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$
2. $\cos ^{-1}(x)=\sin ^{-1}\left(\sqrt{1-x^2}\right)=\tan ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=\cot ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$
3. $\tan ^{-1}(x)=\sin ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)=\cos ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)=\sec ^{-1}\left(\sqrt{1+x^2}\right)=\operatorname{cosec}^{-1}\left(\frac{\sqrt{1+x^2}}{x}\right)$

Approach to Solve Questions of Inverse Trigonometric Functions Class 12

Here are some approaches that students can follow to solve these problems smoothly.

  • Understand the domain and range: Learn the domain and range (Principal value branches) of inverse trigonometric functions efficiently to solve problems.
  • Graphical representation: Draw graphs of inverse trigonometric functions whenever you get stuck to visualise domains and ranges. It also helps to understand the behaviour of the function.
  • Focus on equations: Whenever you get an equation like $\sin ^{-1}(x)=\theta$, convert to $x=\sin (\theta)$ to find exact values.
  • For identity based questions: Use basic identities such as $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}, \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$, to solve composite expressions.
  • Avoid confusion about inverses: Do not confuse $\sin ^{-1} x$ with $(\sin x)^{-1}$. $(\sin x)^{-1}=\frac{1}{\sin x}$, which has a whole different meaning. Similarly, for other trigonometric functions, avoid the same confusion.
  • Shortcut tricks: For MCQ-based questions, put values like $x=\frac12,1,0$ into options to check the answer and cancel out extreme options.

What Extra Should Students Study Beyond the NCERT for JEE?

Here is a comparison list of the concepts in Inverse Trigonometric Functions that are covered in JEE and NCERT, to help students understand what extra they need to study beyond the NCERT for JEE:

NCERT Solutions for Class 12 Maths: Chapter Wise

Given below is the chapter-wise list of the NCERT Class 12 Maths solutions with their respective links:

Also read,

NCERT Solutions class-wise

Given below are the links to class-wise NCERT solutions:

NCERT Books and NCERT Syllabus

Here are the links to NCERT Books and NCERT Syllabus:

Frequently Asked Questions (FAQs)

Q: What are the important topics covered in NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions?
A:

The important topics covered in the Inverse Trigonometric Functions Class 12 Chapter 2 are:

  • Domain and Range: The domain and range for each inverse trigonometric function are explained to help students understand the constraints on values
  • Graphical Representation: How to sketch the graphs of inverse trigonometric functions for better visual understanding.
  • Solving Equations: Techniques to solve equations involving inverse trigonometric functions.
  • Applications: Problems involving real-life applications, including trigonometric equations and angle calculations.
Q: How many exercises are there in NCERT Class 12 Maths Chapter 2
A:

There are 3 exercises in the NCERT class 12 maths chapter 2, they are:

  • Inverse Trigonometric Functions Class 12 Exercise 2.1
  • Inverse Trigonometric Functions Class 12 Exercise 2.2
  • Inverse Trigonometric Functions Class 12 Miscellaneous Exercise
Q: How to simplify inverse trigonometric expressions in Class 12 Maths?
A:

Apply trigonometric identities to express inverse trigonometric functions in simpler forms. Also, use the principal values of the inverse trigonometric functions and convert them to algebraic form whenever necessary or use the substitution method to easily simplify inverse trigonometric expressions in Class 12 Maths.

Q: What are the applications of inverse trigonometric functions in real life?
A:

The main applications of inverse trigonometric functions in real life are:

  • Navigation: Inverse trigonometric functions are used in calculating angles of elevation and depression, as well as directions.
  • Engineering: Used in signal processing, wave analysis, and electrical engineering.
  • Physics: To calculate angles in problems involving vectors, motion, and forces.
  • Architecture: Helps in designing structures by calculating angles and slopes.
Q: How to prove standard properties of inverse trigonometric functions?
A:

To prove standard properties of inverse trigonometric functions, you can use definitions of Inverse Trigonometric Functions, Algebraic manipulations, Trigonometric Identities, Right Triangle approach, Graphs (for Principal Values), and the Substitution method.

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Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?

Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.



Hello

For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.

Hello,

If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

For the 2025-2026 academic session, the CBSE plans to conduct board exams from 17 February 2026 to 20 May 2026.

You can download it in pdf form from below link

CBSE DATE SHEET 2026

all the best for your exam!!

Hii neeraj!

You can check CBSE class 12th registration number in:

  • Your class 12th board exam admit card. Please do check admit card for registration number, it must be there.
  • You can also check the registration number in your class 12th marksheet in case you have got it.
  • Alternatively you can also visit your school and ask for the same in the administration office they may tell you the registration number.

Hope it helps!