Class 12 Inverse Trig Functions Exercise

Q: Q: 1. What is a circular function?

Ans: We know that \sin \theta, \cos\theta, \tan\theta etc. are called trigonometrical circular functions, each of which is equal to a certain number for each of them.

Q: Q: 2. What are inverse circular functions?

Ans: If \sin\theta = x then \theta = \sin^{-1}x . The expression that expresses the angle \theta as x is called the inverse circular function. Similarly, the other inverse function expressing the angle as a number x is \cos^{-1}x, \tan^{-1}x, \csc^{-1}x, \sec^{-1}x and \cot^{-1}x.

Q: Q: 3. Write a brief note on the principal value.

Ans: We know that general value of \sin \theta =\sin \{ n\pi + (-1)^n \theta \} Where n \in Z is set of integers. Where if \sin^{-1} x = \theta, then the general value of \sin^{-1} x, n\pi + (-1)^n \sin^{-1} x is called \text{Sin}^{-1} x denoted by \text{Sin}^{-1} x = n\pi + (-1)^n \sin^{-1} x, n \in \mathbb{Z}.

Sanjay Kumawat November 8, 2026 12th Mathematics(English) No Comments

Contents hide

1 1.Class 12 Inverse Trig Functions Exercise

2 2.Class 12 Inverse Trig Functions:

2.1 3.Practice Questions of Class 12 Inverse Trig Functions Exercise for Students

2.2 4.Frequently Asked Questions Related to Class 12 Inverse Trig Functions Exercise

2.3 Q: 1. What is a circular function?

2.4 Q: 2. What are inverse circular functions?

2.5 Q: 3. Write a brief note on the principal value.

2.5.1 5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

1.Class 12 Inverse Trig Functions Exercise

Class 12 Inverse Trig Functions Exercise

Learn,”Class 12 Inverse Trig Functions Exercise”,Exercise Based questions,simplest form and other specific questions.

Also Read This Article:- Principal Value: Inverse Trig Class 12

2.Class 12 Inverse Trig Functions:

Prove the following:
Example: 1. $3 \sin^{-1} x = \sin^{-1}(3x - 4x^3), \quad x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$
Solution: Let $\sin^{-1} x = \theta \implies \sin \theta = x$
$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$
$= 3x - 4x^3$
$\implies 3\theta = \sin^{-1}(3x - 4x^3)$
$\implies 3 \sin^{-1} x = \sin^{-1}(3x - 4x^3)$
Example: 2. $3 \cos^{-1} x = \cos^{-1}(4x^3 - 3x), \quad x \in \left[\frac{1}{2}, 1\right]$
Solution: $3 \cos^{-1} x = \cos^{-1}(4x^3 - 3x)$
Let $\cos^{-1} x = \theta \implies x = \cos \theta$
$\cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta$
$3\theta = \cos^{-1}(4 \cos^3 \theta - 3 \cos \theta)$
$= \cos^{-1}(4x^3 - 3x)$
$\implies 3 \cos^{-1} x = \cos^{-1}(4x^3 - 3x)$
Write the following function in the simplest form:
Example: 3. $\tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right), \quad x \neq 0$
Solution: $\tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right)$
Put $x = \tan \theta \implies \theta = \tan^{-1} x$
$\tan^{-1}\left(\frac{\sqrt{1 + \tan^2 \theta} - 1}{\tan \theta}\right)$
$= \tan^{-1}\left(\frac{\sqrt{\sec^2 \theta }- 1}{\tan \theta}\right)$
$= \tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right)$
$= \tan^{-1}\left(\frac{\frac{1 - \cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}\right)$
$= \tan^{-1}\left(\frac{1 - \cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\right)$
$= \tan^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right)$
$= \tan^{-1}\left[\frac{1 - (1 - 2\sin^2\frac{\theta}{2})}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right]$
$= \tan^{-1}\left[\frac{1 - 1 + 2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right]$
$= \tan^{-1}\left[\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right]$
$= \tan^{-1}\left[\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\right]$
$= \tan^{-1}\left(\tan\frac{\theta}{2}\right)$
$= \frac{\theta}{2} = \frac{1}{2} \tan^{-1} x$
Example: 4. $\tan^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right) , 0 < x < \pi$
Solution: $\tan^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right)$
$= \tan^{-1}\left(\sqrt{\frac{1 - (1 - 2\sin^2\frac{x}{2})}{1 + 2\cos^2\frac{x}{2} - 1}}\right)$
$= \tan^{-1}\left(\sqrt{\frac{1 - 1 + 2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}\right)$
$= \tan^{-1}\left(\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}\right)$
$= \tan^{-1}\left[\tan\frac{x}{2}\right]$
$= \frac{x}{2}$
Example: 5. $\tan^{-1}\left(\frac{\sin x - \cos x}{\cos x + \sin x}\right) , -\frac{\pi}{4} < x<\frac{3 \pi}{4}$
Solution: $\tan^{-1}\left(\frac{\sin x - \cos x}{\cos x + \sin x}\right)$
$= \tan^{-1}\left[\frac{\frac{\sin x}{\cos x} - 1}{1 + \frac{\sin x}{\cos x}}\right]$
$= \tan^{-1}\left[\frac{\tan x - \tan\frac{\pi}{4}}{1 + \tan\frac{\pi}{4}\tan x}\right]$
$= \tan^{-1}\left[\tan\left(x - \frac{\pi}{4}\right)\right]$
$= x - \frac{\pi}{4}$
Example: 6. $\tan^{-1}\left(\frac{x}{\sqrt{a^2 - x^2}}\right), \quad |x| < a$
Solution: $\tan^{-1}\left(\frac{x}{\sqrt{a^2 - x^2}}\right)$
Put $x = a \sin \theta \implies \theta = \sin^{-1}\left(\frac{x}{a}\right)$
$= \tan^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2 - a^2 \sin^2 \theta}}\right)$
$= \tan^{-1}\left(\frac{a \sin \theta}{a \sqrt{1 - \sin^2 \theta}}\right)$
$= \tan^{-1}\left(\frac{\sin \theta}{\sqrt{\cos^2 \theta}}\right)$
$= \tan^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)$
$= \tan^{-1}(\tan \theta)$
$= \theta$
$= \sin^{-1}\left(\frac{x}{a}\right)$
Example: 7. $\tan^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}\right), \quad a > 0; \quad -\frac{a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$
Solution: $\tan^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}\right)$
Put $x = a \tan \theta \implies \theta = \tan^{-1}\left(\frac{x}{a}\right)$
$\tan^{-1}\left(\frac{3a^3 \tan \theta - a^3 \tan^3 \theta}{a^3 - 3a^3 \tan^2 \theta}\right)$
$= \tan^{-1}\left[\frac{a^3(3 \tan \theta - \tan^3 \theta)}{a^3(1 - 3 \tan^2 \theta)}\right]$
$= \tan^{-1}\left(\frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}\right)$
$= \tan^{-1} \tan(3\theta)$
$= 3\theta$
$= 3 \tan^{-1}\left(\frac{x}{a}\right)$
Find the values of each of the following:
Example: 8. $\tan^{-1}\left[2 \cos\left(2 \sin^{-1} \frac{1}{2}\right)\right]$
Solution: $\tan^{-1}\left[2 \cos\left(2 \sin^{-1} \frac{1}{2}\right)\right]$
$= \tan^{-1}\left[2 \cos\left(\sin^{-1}\left(2 \times \frac{1}{2} \sqrt{1 - \frac{1}{4}}\right)\right)\right] \quad [\because 2\sin^{-1}x = \sin^{-1}(2x\sqrt{1-x^2})]$
$= \tan^{-1}\left[2 \cos\left(\sin^{-1} \frac{\sqrt{3}}{2}\right)\right]$
$= \tan^{-1}\left[2 \cos\left(\frac{\pi}{3}\right)\right]$
$= \tan^{-1}\left[2 \times \frac{1}{2}\right]$
$= \tan^{-1}(1)$
$= \frac{\pi}{4}$

Class 12 Inverse Trig Functions Exercise

Example: 9. $\tan \frac{1}{2} \left[\sin^{-1}\left(\frac{2x}{1 + x^2}\right) + \cos^{-1}\left(\frac{1 - y^2}{1 + y^2}\right)\right], \quad |x| < 1, y > 0 \text{ and } xy < 1$
Solution: $\tan \frac{1}{2} \left[\sin^{-1}\left(\frac{2x}{1 + x^2}\right) + \cos^{-1}\left(\frac{1 - y^2}{1 + y^2}\right)\right]$
Put $x = \tan \theta_1$ and $y = \tan \theta_2$
$\theta_1 = \tan^{-1} x$ and $\theta_2 = \tan^{-1} y$
$= \tan \frac{1}{2} \left[\sin^{-1}\left(\frac{2\tan \theta_1}{1 + \tan^2 \theta_1}\right) + \cos^{-1}\left(\frac{1 - \tan^2 \theta_2}{1 + \tan^2 \theta_2}\right)\right]$
$= \tan \frac{1}{2} [\sin^{-1} \sin(2\theta_1) + \cos^{-1} \cos(2\theta_2)]$
$= \tan \frac{1}{2} [2\theta_1 + 2\theta_2]$
$= \tan \left[\frac{1}{2} \times 2(\theta_1 + \theta_2)\right]$
$= \tan(\theta_1 + \theta_2)$
$= \tan(\tan^{-1} x + \tan^{-1} y)$
$= \tan\left[\tan^{-1}\left(\frac{x + y}{1 - xy}\right)\right]$
$= \frac{x + y}{1 - xy}$
Find the values of each of the expressions in Exercises 10 to 15:
Example: 10. $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$
Solution: $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$
$= \sin^{-1}\left[\sin\left(\pi - \frac{\pi}{3}\right)\right]$
$= \sin^{-1}\left[\sin\frac{\pi}{3}\right]$
$= \frac{\pi}{3}$
Example: 11. $\tan^{-1}\left(\tan\frac{3\pi}{4}\right)$
Solution: $\tan^{-1}\left(\tan\frac{3\pi}{4}\right)$
$= \tan^{-1}\left(\tan\left(\pi - \frac{\pi}{4}\right)\right)$
$= \tan^{-1}\left(-\tan\frac{\pi}{4}\right) = -\frac{\pi}{4}$
Example: 12. $\tan\left(\sin^{-1}\frac{3}{5} + \cot^{-1}\frac{3}{2}\right)$
Solution: $\tan\left(\sin^{-1}\frac{3}{5} + \cot^{-1}\frac{3}{2}\right)$
$= \tan\left[\tan^{-1}\frac{3}{\sqrt{5^2 - 3^2}} + \tan^{-1}\frac{2}{3}\right]$
$= \tan\left[\tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\frac{2}{3}\right]$
$= \tan\left[\tan^{-1}\left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}}\right)\right]$
$= \frac{\frac{9 + 8}{12}}{1 - \frac{1}{2}} = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{12} \times \frac{2}{1}$
$= \frac{17}{6}$
Example: 13. $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$ is equal to:
(A) $\frac{7\pi}{6}$ (B) $\frac{5\pi}{6}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{6}$
Solution: $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$
$= \cos^{-1}\left[\cos\left(\pi + \frac{\pi}{6}\right)\right]$
$= \cos^{-1}\left(-\cos\frac{\pi}{6}\right)$
$= \pi - \cos^{-1}\left(\cos\frac{\pi}{6}\right)$
$= \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
विकल्प (B) सही है
Example: 14. $\sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to
(A) $\frac{1}{2}$ (B) $\frac{1}{3}$ (C) $\frac{1}{4}$ (D) $1$
Solution: $\sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right)$
$= \sin\left(\frac{\pi}{3} + \sin^{-1}\left(\frac{1}{2}\right)\right)$
$= \sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right)$
$= \sin\left(\frac{2\pi + \pi}{6}\right) = \sin\left(\frac{3\pi}{6}\right)$
$= \sin\left(\frac{\pi}{2}\right) = 1$
विकल्प (D) सही है
Example: 15. $\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})$ is equal to
(A) $\pi$ (B) $-\frac{\pi}{2}$ (C) $0$ (D) $2\sqrt{3}$
Solution: $\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})$
$= \tan^{-1}\sqrt{3} - (\pi - \cot^{-1}\sqrt{3})$
$= \tan^{-1}\sqrt{3} - \pi + \cot^{-1}(\sqrt{3})$
$= \tan^{-1}\sqrt{3} - \pi + \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$= \frac{\pi}{3} - \pi + \frac{\pi}{6}$
$= \frac{2\pi - 6\pi + \pi}{6}$
$= -\frac{3\pi}{6}$
$= -\frac{\pi}{2}$
विकल्प (B) सही है।

Also Read This Article:- Inverse trigonometric functions

3.Practice Questions of Class 12 Inverse Trig Functions Exercise for Students

Class 12 Inverse Trig Functions Exercise

(1.) Prove that
$\tan\left(\frac{\pi}{4} + \frac{1}{2} \cos^{-1}\left(\frac{a}{b}\right)\right) + \tan\left(\frac{\pi}{4} - \frac{1}{2} \cos^{-1}\left(\frac{a}{b}\right)\right) = \frac{2b}{a}$
(2.) Prove that
$\cos^{-1}\left(\frac{1 - a^2}{1 + a^2}\right) + \cos^{-1}\left(\frac{1 - b^2}{1 + b^2}\right) = 2 \tan^{-1}(x)$
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Also Read This Article:- Properties of Inverse Trig Functions Class 12

4.Frequently Asked Questions Related to Class 12 Inverse Trig Functions Exercise

Q: 1. What is a circular function?

Ans: We know that $\sin \theta, \cos\theta, \tan\theta$ etc. are called trigonometrical circular functions, each of which is equal to a certain number for each of them.

Q: 2. What are inverse circular functions?

Ans: If $\sin\theta = x$ then $\theta = \sin^{-1}x$ . The expression that expresses the angle $\theta$ as x is called the inverse circular function. Similarly, the other inverse function expressing the angle as a number x is $\cos^{-1}x, \tan^{-1}x, \csc^{-1}x, \sec^{-1}x$ and $\cot^{-1}x$ .

Q: 3. Write a brief note on the principal value.

Ans: We know that general value of $\sin \theta =\sin \{ n\pi + (-1)^n \theta \}$
Where $n \in Z$ is set of integers.
Where if $\sin^{-1} x = \theta$ , then the general value of $\sin^{-1} x, n\pi + (-1)^n \sin^{-1} x$ is called $\text{Sin}^{-1} x$ denoted by $\text{Sin}^{-1} x = n\pi + (-1)^n \sin^{-1} x, n \in \mathbb{Z}$ .

5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

Class 12 Inverse Trig Functions Exercise

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अगर $\pi = \frac{22}{7}$ और r = 3.5, तो $2\pi r^2$ का मान क्या होगा?
(If $\pi = \frac{22}{7}$ and r = 3.5, then what is the value of $2\pi r^2$ ?)
*पिछली प्रश्नोत्तरी का हल*
$5 * 9 = \sqrt{\frac{(5 + 9)(9^2 - 12 \times 5)}{(5 - 2)(9 - 7)}}$
$= \sqrt{\frac{14 \times 21}{3 \times 2}} = 7$
*Previous Quiz Solution*
$5 * 9 = \sqrt{\frac{(5 + 9)(9^2 - 12 \times 5)}{(5 - 2)(9 - 7)}}$
$= \sqrt{\frac{14 \times 21}{3 \times 2}} = 7$

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