NCERT Solutions for Class 12 Maths Chapter 2

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

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

NCERT Solutions for Class 12 Maths Chapter 2 – Inverse Trigonometric Functions helps students understand principal values, domain, and range effectively. Moreover, structured solutions improve accuracy and confidence. Therefore, regular practice strengthens conceptual clarity and prepares students for board exams with better performance and problem solving skills.

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Students can access NCERT Solutions for Class 12 Maths Chapter 2 – Inverse Trigonometric Functions PDF for organized revision and practice. Moreover, downloadable resources help in offline learning. Therefore, consistent revision improves accuracy and strengthens conceptual understanding for board exam preparation and academic success.

NCERT – Exercise – 2.1

Find the principal values of the following:

  1. ${\sin ^{ – 1}}\left( { – \frac{1}{2}} \right)$

    SOLUTION

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

    We know that the range of the principal value branch of ${\sin ^{ – 1}}$ is $\left[ { – \frac{\pi }{2},\;\;\frac{\pi }{2}} \right].$

    Then, $\sin \left( { – \frac{\pi }{6}} \right) = – \frac{1}{2},\,\;where\;\; – \frac{\pi }{6} \in \left[ { – \frac{\pi }{2},\;\;\frac{\pi }{2}} \right]$

    Hence, the principal value of ${\sin ^{ – 1}}\left( { – \frac{1}{2}} \right)\;\;is\;\; – \frac{\pi }{6}$

  2. ${\cos ^{ – 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)$

    SOLUTION

    Let ${\cos ^{ – 1}}\frac{{\sqrt 3 }}{2} = x \Rightarrow \frac{{\sqrt 3 }}{2} = \cos x$

    We know that the range of the principal value branch of ${\cos ^{ – 1}}$ is $[0,\;\pi ]$

    Then, $\cos x = \frac{{\sqrt 3 }}{2} = \cos \frac{\pi }{6},\;\;where\;\;\frac{\pi }{6} \in [0,\;\pi ]$

    Hence, the principal value of ${\cos ^{ – 1}}\frac{{\sqrt 3 }}{2}$ is $\frac{\pi }{6}.$

  3. ${\rm{cose}}{{\rm{c}}^{ – 1}}(2)$

    SOLUTION

    Let ${\rm{cose}}{{\rm{c}}^{ – 1}}(2) = x \Rightarrow 2 = {\rm{cosec}}\;{\rm{x}}$

    We know that the range of the principal value branch of ${\rm{cose}}{{\rm{c}}^{ – 1}}\;\;is\;\;\left[ { – \frac{\pi }{2},\;\frac{\pi }{2}} \right] – \{ 0\} $

    Then, $2 = {\rm{cosec}}\;{\rm{x}}\;{\rm{ = cosec }}\left( {\frac{\pi }{6}} \right),\;where\;\frac{\pi }{6} \in \left[ {\frac{{ – \pi }}{2},\;\frac{\pi }{2}} \right] – \{ 0\} $

    Hence, the principal value of ${\rm{cose}}{{\rm{c}}^{ – 1}}(2)\;is\;\pi /6.$

  4. ${\tan ^{ – 1}}( – \sqrt 3 )$

    SOLUTION

    Let ${\tan ^{ – 1}}( – \sqrt 3 ) = x \Rightarrow – \sqrt 3 = \tan x$

    We know that the range of the principal value branch of ${\tan ^{ – 1}}$ is $\left( { – \frac{\pi }{2},\;\frac{\pi }{2} } \right)$

    Then, $\tan x = – \sqrt 3 = \tan \left( { – \frac{\pi }{3}} \right),\;\;where – \frac{\pi }{3} \in \left( { – \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

    Hence, the principal value of ${\tan ^{ – 1}}( – \sqrt 3 )$ is $ – \frac{\pi }{3}.$

  5. ${\cos ^{ – 1}}\left( { – \frac{1}{2}} \right)$

    SOLUTION

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

    We know that the range of principal value branch of ${\cos ^{ – 1}}\;is\;[0,\;\pi ]$

    Then, $\left( { – \frac{1}{2}} \right) = – \cos \frac{\pi }{3} = \cos \left( {\pi – \frac{\pi }{3}} \right) = \cos \left( {\frac{{2\pi }}{3}} \right),$

    Where $\frac{{2\pi }}{3} \in [0,\;\pi ]$

    Hence, the principal value of ${\cos ^{ – 1}}\left( { – \frac{1}{2}} \right)\;is\;\frac{{2\pi }}{3}.$

  6. ${\tan ^{ – 1}}( – 1)$

    SOLUTION

    Let ${\tan ^{ – 1}}( – 1) = x \Rightarrow – 1 = \tan x$

    We know that the range of principal value branch of ${\tan ^{ – 1}}$ is $\left( { – \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

    Then, $ – 1 = \tan \left( { – \frac{\pi }{4}} \right)where – \frac{\pi }{4} \in \left( { – \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

    Hence, the principal value of ${\tan ^{ – 1}}( – 1)\;is\; – \frac{\pi }{4}.$

  7. ${\sec ^{ – 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)$

    SOLUTION

    Let ${\sec ^{ – 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) = x \Rightarrow \sec x = \frac{2}{{\sqrt 3 }}$

    We know that the range of principal value branch of ${\sec ^{ – 1}}$ is $[0,\;\pi ] – \left\{ {\frac{\pi }{2}} \right\}.$

    Then, $\frac{2}{{\sqrt 3 }} = \sec \left( {\frac{\pi }{6}} \right),\;\;where\;\frac{\pi }{6} \in [0,\;\pi ] – \left\{ {\frac{\pi }{2}} \right\}$

    Hence, the principal value of ${\sec ^{ – 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)\;\,is\;\;\frac{\pi }{6}.$

  8. ${\cot ^{ – 1}}(\sqrt 3 )$

    SOLUTION

    Let ${\cot ^{ – 1}}(\sqrt 3 ) = x \Rightarrow \sqrt 3 = \cot x$

    We know that the range of principal value of ${\cot ^{ – 1}}\;is\;(0,\;\pi )$

    Then, $\sqrt 3 = \cot \left( {\frac{\pi }{6}} \right),\;\;where\;\frac{\pi }{6} \in (0,\;\pi )$

    Hence, the principal value branch of ${\cot ^{ – 1}}(\sqrt 3 )\;\;is\;\frac{\pi }{6}.$

  9. ${\cos ^{ – 1}}\left( { – \frac{1}{{\sqrt 2 }}} \right)$

    SOLUTION

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

    We know that the range of principal value branch of ${\cos ^{ – 1}}is\;[0,\;\pi ]$

    Then, $ – \frac{1}{{\sqrt 2 }} = – \cos \left( {\frac{\pi }{4}} \right) = \cos \left( {\pi – \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4},$

    Where $\frac{{3\pi }}{4} \in [0,\;\pi ]$

    Hence, the principal value of ${\cos ^{ – 1}}\left( { – \frac{1}{{\sqrt 2 }}} \right)$ is $\frac{{3\pi }}{4}.$

  10. ${\rm{cose}}{{\rm{c}}^{ – 1}}( – \sqrt 2 )$

    SOLUTION

    Let ${\rm{cose}}{{\rm{c}}^{ – 1}}( – \sqrt 2 ) = x \Rightarrow – \sqrt 2 = {\rm{cosec}}\;{\rm{x}}$

    We know that the range of principal value branch of ${\rm{cose}}{{\rm{c}}^{ – 1}}$ is $\left[ { – \frac{\pi }{2},\;\frac{\pi }{2}} \right] – \{ 0\} $

    Then, $ – \sqrt 2 = – {\rm{cosec}}\left( {\frac{\pi }{4}} \right) = {\rm{cosec}}\left( { – \frac{\pi }{4}} \right),$

    Where $ – \frac{\pi }{4} \in \left[ { – \frac{\pi }{2},\;\frac{\pi }{2}} \right] – \{ 0\} $

    Hence, the principal value of ${\rm{cose}}{{\rm{c}}^{ – 1}}( – \sqrt 2 )\;is\; – \frac{\pi }{4}.$

    11. Find the values of the following:

  11. ${\tan ^{ – 1}}(1) + {\cos ^{ – 1}}\left( { – \frac{1}{2}} \right) + {\sin ^{ – 1}}\left( { – \frac{1}{2}} \right)$

    SOLUTION

    We know that the range of principal value branch of ${\tan ^{ – 1}},\;\;{\cos ^{ – 1}}\;\;and\;\;{\sin ^{ – 1}}$ are $\left( { – \frac{\pi }{2},\;\,\frac{\pi }{2}} \right),\;[0,\;\pi ]\;$ and $\left[ { – \frac{\pi }{2},\;\frac{\pi }{2}} \right]$ respectively.

    Let ${\tan ^{ – 1}}(1) = x \Rightarrow 1 = \tan x$ Then, $1 = \tan \left( {\frac{\pi }{4}} \right),\;\,where\;\;\frac{\pi }{4} \in \left( { – \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

    Let ${\cos ^{ – 1}}\left( { – \frac{1}{2}} \right) = y \Rightarrow – \frac{1}{2} = \cos y$ Then, $ – \frac{1}{2} = – \cos \left( {\frac{\pi }{3}} \right) = \cos \left( {\pi – \frac{\pi }{3}} \right) = \cos \frac{{2\pi }}{3},$

    Where $\frac{{2\pi }}{3} \in [0,\;\pi ]$

    Let ${\sin ^{ – 1}}\left( { – \frac{1}{2}} \right) = z \Rightarrow – \frac{1}{2} = \sin z$ Then, $ – \frac{1}{2} = \sin \left( { – \frac{\pi }{6}} \right), where – \frac{\pi }{6} \in \left[ { – \frac{\pi }{2},\;\frac{\pi }{2}} \right]$

    So, ${\tan ^{ – 1}}(1) + {\cos ^{ – 1}}\left( { – \frac{1}{2}} \right) + {\sin ^{ – 1}}\left( { – \frac{1}{2}} \right) = \left( {\frac{\pi }{4} + \frac{{2\pi }}{3} – \frac{\pi }{6}} \right) = \frac{{3\pi }}{4}.$

  12. ${\cos ^{ – 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ – 1}}\left( {\frac{1}{2}} \right)$

    SOLUTION

    We know that the range of principal value branch of ${\cos ^{ – 1}}$ and ${\sin ^{ – 1}}$ are $[0,\;\pi ]$ and $\left[ { – \frac{\pi }{2},\;\frac{\pi }{2}} \right]$ respectively.

    Let ${\cos ^{ – 1}}\left( {\frac{1}{2}} \right) = x \Rightarrow \frac{1}{2} = \cos x$ Then, $\frac{1}{2} = \cos \left( {\frac{\pi }{3}} \right),\;where\;\frac{\pi }{3} \in [0,\;\pi ]$

    Let ${\sin ^{ – 1}}\left( {\frac{1}{2}} \right) = y \Rightarrow \frac{1}{2} = \sin y$ Then, $\frac{1}{2} = \sin \left( {\frac{\pi }{6}} \right),\;\;where\;\frac{\pi }{6} \in \left[ { – \frac{\pi }{2},\;\frac{\pi }{2}} \right]$

    $\therefore$ ${\cos ^{ – 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ – 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{3} + 2 \cdot \frac{\pi }{6} = \frac{\pi }{3} + \frac{\pi }{3} = \frac{{2\pi }}{3}.$

  13. If ${\sin ^{ – 1}}x = y,$ then

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

    (B) $ – \frac{\pi }{2} \le y \le \frac{\pi }{2}$

    (C) $0 < y < \pi $

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

    SOLUTION

    (B) ${\sin ^{ – 1}}x = y$ $ \Rightarrow $ $x = \sin y,$ where the range of principal value branch of ${\sin ^{ – 1}}\;is\;\left[ { – \frac{\pi }{2},\;\frac{\pi }{2}} \right],\;then\;y \in \left[ { – \frac{\pi }{2},\;\frac{\pi }{2}} \right] \Rightarrow – \frac{\pi }{2} \le y \le \frac{\pi }{2}.$

  14. ${\tan ^{ – 1}}\sqrt 3 – {\sec ^{ – 1}}( – 2)$ is equal to

    (A) $\pi $

    (B) $ – \frac{\pi }{3}$

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

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

    SOLUTION

    (B) We know that the range of principal value branch of ${\tan ^{ – 1}}$ and ${\sec ^{ – 1}}$ are $\left( { – \frac{\pi }{2},\;\frac{\pi }{2}} \right)$ and $[0,\;\pi ] – \left\{ {\frac{\pi }{2}} \right\}$ respectively.

    Let ${\tan ^{ – 1}}(\sqrt 3 ) = x \Rightarrow \sqrt 3 = \tan x,\;then,\;\sqrt 3 = \tan \left( {\frac{\pi }{3}} \right),\;where \frac{\pi }{3} \in \left( { – \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

    Let ${\sec ^{ – 1}}( – 2) = y \Rightarrow – 2 = \sec y$ Then, $ – 2 = – \sec \left( {\frac{\pi }{3}} \right) = \sec \left( {\pi – \frac{\pi }{3}} \right) = \sec \frac{{2\pi }}{3},$

    Where $\frac{{2\pi }}{3} \in [0,\;\pi ] – \left\{ {\frac{\pi }{2}} \right\}$

    $\therefore$ ${\tan ^{ – 1}}\sqrt 3 – {\sec ^{ – 1}}( – 2) = \frac{\pi }{3} – \frac{{2\pi }}{3} = \frac{\pi }{3}.$

NCERT – Exercise – 2.2

Prove the Following:

1. $3\sin^{-1}x = \sin^{-1}(3x – 4x^3),\;x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$
Solution:

Let $\sin^{-1}x = \theta$. Then, $x = \sin \theta$.
Now, $\sin 3\theta = 3\sin \theta – 4\sin^3 \theta = 3x – 4x^3$.
Therefore, $3\theta = \sin^{-1}(3x – 4x^3)$.
Hence, $3\sin^{-1}x = \sin^{-1}(3x – 4x^3)$.

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

Let $\cos^{-1}x = \theta$. Then, $x = \cos \theta$.
Now, $\cos 3\theta = 4\cos^3 \theta – 3\cos \theta = 4x^3 – 3x$.
Therefore, $3\theta = \cos^{-1}(4x^3 – 3x)$.
Hence, $3\cos^{-1}x = \cos^{-1}(4x^3 – 3x)$.

3. $\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24} = \tan^{-1}\frac{1}{2}$
Solution:

$\tan^{-1}\left(\frac{2}{11}\right) + \tan^{-1}\left(\frac{7}{24}\right)$
$= \tan^{-1}\left[\frac{\frac{2}{11} + \frac{7}{24}}{1 – \frac{2}{11} \times \frac{7}{24}}\right]$
$= \tan^{-1}\left[\frac{\frac{48 + 77}{264}}{\frac{264 – 14}{264}}\right]$
$= \tan^{-1}\left[\frac{125}{250}\right] = \tan^{-1}\left(\frac{1}{2}\right)$.

4. $2\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}$
Solution:

$2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right)$
$= \tan^{-1}\frac{2 \times \frac{1}{2}}{1 – \left(\frac{1}{2}\right)^2} + \tan^{-1}\left(\frac{1}{7}\right)$
$= \tan^{-1}\left(\frac{1}{\frac{3}{4}}\right) + \tan^{-1}\left(\frac{1}{7}\right)$
$= \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right)$
$= \tan^{-1}\left[\frac{\frac{4}{3} + \frac{1}{7}}{1 – \frac{4}{3} \times \frac{1}{7}}\right]$
$= \tan^{-1}\left(\frac{\frac{28 + 3}{21}}{\frac{21 – 4}{21}}\right)$
$= \tan^{-1}\left(\frac{31}{17}\right)$.

Write the following functions in the simplest form:

5. $\tan^{-1}\frac{\sqrt{1 + x^2} – 1}{x},\;x \neq 0$
Solution:

Let $x = \tan \theta$.
$\tan^{-1}\left(\frac{\sqrt{1 + x^2} – 1}{x}\right)$
$= \tan^{-1}\left[\frac{\sqrt{1 + \tan^2 \theta} – 1}{\tan \theta}\right]$
$= \tan^{-1}\left(\frac{\sec \theta – 1}{\tan \theta}\right)$
$= \tan^{-1}\left(\frac{1 – \cos \theta}{\sin \theta}\right)$
$= \tan^{-1}\left[\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right]$
$= \tan^{-1}\left(\tan \frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2}\tan^{-1}x$.

6. $\tan^{-1}\frac{1}{\sqrt{x^2 – 1}},\;|x| > 1$
Solution:

Let $x = \sec \theta$.
$\tan^{-1}\left(\frac{1}{\sqrt{x^2 – 1}}\right)$
$= \tan^{-1}\left(\frac{1}{\sqrt{\sec^2 \theta – 1}}\right)$
$= \tan^{-1}\left(\frac{1}{\tan \theta}\right)$
$= \tan^{-1}(\cot \theta)$
$= \tan^{-1}\left\{\tan\left(\frac{\pi}{2} – \theta\right)\right\}$
$= \frac{\pi}{2} – \theta = \frac{\pi}{2} – \sec^{-1}x$.

7. $\tan^{-1}\left(\sqrt{\frac{1 – \cos x}{1 + \cos x}}\right),\;x < \pi$
Solution:

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

8. $\tan^{-1}\left(\frac{\cos x – \sin x}{\cos x + \sin x}\right),\;0 < x < \pi$
Solution:

$\tan^{-1}\left(\frac{\cos x – \sin x}{\cos x + \sin x}\right)$
$= \tan^{-1}\left(\frac{1 – \tan x}{1 + \tan x}\right)$
$= \tan^{-1}\left(\tan\left(\frac{\pi}{4} – x\right)\right)$
$= \frac{\pi}{4} – x$.

9. $\tan^{-1}\left(\frac{x}{\sqrt{a^2 – x^2}}\right),\;|x| < a$
Solution:

Let $x = a \sin \theta$.
$\tan^{-1}\left(\frac{x}{\sqrt{a^2 – x^2}}\right)$
$= \tan^{-1}\left(\frac{a \sin \theta}{a \cos \theta}\right)$
$= \tan^{-1}(\tan \theta) = \theta = \sin^{-1}\left(\frac{x}{a}\right)$.

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

Let $x = a \tan \theta$.
$\tan^{-1}\left(\frac{3a^2x – x^3}{a^3 – 3ax^2}\right)$
$= \tan^{-1}\left[\frac{3a^3 \tan \theta – a^3 \tan^3 \theta}{a^3 – 3a^3 \tan^2 \theta}\right]$
$= \tan^{-1}\left(\frac{3 \tan \theta – \tan^3 \theta}{1 – 3 \tan^2 \theta}\right)$
$= \tan^{-1}(\tan 3\theta) = 3\theta = 3\tan^{-1}\left(\frac{x}{a}\right)$.

Find the values of each of the following:

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

$\tan^{-1}\left\{2\cos\left(2\sin^{-1}\frac{1}{2}\right)\right\}$
$= \tan^{-1}\left\{2\cos\left(2 \times \frac{\pi}{6}\right)\right\}$
$= \tan^{-1}\left\{2\cos\frac{\pi}{3}\right\}$
$= \tan^{-1}\left[2 \times \frac{1}{2}\right] = \tan^{-1}1 = \frac{\pi}{4}$.

12. $\cot\left(\tan^{-1}a + \cot^{-1}a\right)$
Solution:

$\cot\left(\tan^{-1}a + \cot^{-1}a\right)$
$= \cot\left(\tan^{-1}a + \frac{\pi}{2} – \tan^{-1}a\right)$
$= \cot\frac{\pi}{2} = 0$.

13. $\tan\frac{1}{2}\left[\sin^{-1}\frac{2x}{1 + x^2} + \cos^{-1}\frac{1 – y^2}{1 + y^2}\right],\;|x| < 1,\;y > 0$ and $xy < 1$
Solution:

Let $x = \tan \theta$ and $y = \tan \phi$.
$\tan\left\{\frac{1}{2}\sin^{-1}\left(\frac{2x}{1 + x^2}\right) + \frac{1}{2}\cos^{-1}\left(\frac{1 – y^2}{1 + y^2}\right)\right\}$
$= \tan\left\{\frac{1}{2}\sin^{-1}\left(\frac{2\tan \theta}{1 + \tan^2 \theta}\right) + \frac{1}{2}\cos^{-1}\left(\frac{1 – \tan^2 \phi}{1 + \tan^2 \phi}\right)\right\}$
$= \tan\left\{\frac{1}{2}\sin^{-1}(\sin 2\theta) + \frac{1}{2}\cos^{-1}(\cos 2\phi)\right\}$
$= \tan\left\{\frac{1}{2} \times 2\theta + \frac{1}{2} \times 2\phi\right\}$
$= \tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 – \tan \theta \tan \phi}$
$= \frac{x + y}{1 – xy}$.

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

$\sin\left(\sin^{-1}\frac{1}{5} + \cos^{-1}x\right) = 1$
$\Rightarrow \sin^{-1}\frac{1}{5} + \cos^{-1}x = \sin^{-1}1$
$\Rightarrow \sin^{-1}\frac{1}{5} + \frac{\pi}{2} – \sin^{-1}x = \frac{\pi}{2}$
$\Rightarrow \sin^{-1}\frac{1}{5} = \sin^{-1}x$
$\Rightarrow x = \sin\left(\sin^{-1}\frac{1}{5}\right) = \frac{1}{5}$.

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

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

Find the value of each of the expressions:

16. $\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$
Solution:

$\sin^{-1}\left(\sin\frac{2\pi}{3}\right) \neq \frac{2\pi}{3}$
as the principal value branch of $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
So, $\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \sin^{-1}\left(\sin\left(\pi – \frac{\pi}{3}\right)\right)$
$= \sin^{-1}\left(\sin\frac{\pi}{3}\right) = \frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Hence, $\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{\pi}{3}$.

17. $\tan^{-1}\left(\tan\frac{3\pi}{4}\right)$
Solution:

$\tan^{-1}\left(\tan\frac{3\pi}{4}\right) \neq \frac{3\pi}{4}$
as the principal value branch of $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
So, $\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \tan^{-1}\left(\tan\left(\pi – \frac{\pi}{4}\right)\right)$
$= \tan^{-1}\left(-\tan\frac{\pi}{4}\right) = \tan^{-1}\left(\tan\left(-\frac{\pi}{4}\right)\right)$
$= -\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Hence, $\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = -\frac{\pi}{4}$.

18. $\tan\left[\left(\sin^{-1}\frac{3}{5}\right) + \cot^{-1}\frac{3}{2}\right]$
Solution:

Let $A = \sin^{-1}\frac{3}{5}$ and $B = \cot^{-1}\frac{3}{2}$.
$\Rightarrow \sin A = \frac{3}{5}$ and $\cot B = \frac{3}{2}$
$\Rightarrow \tan A = \frac{3}{4}$ and $\tan B = \frac{2}{3}$
So, $\tan\left(\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{2}{3}\right)$
$= \tan\left[\tan^{-1}\left(\frac{\frac{3}{4} + \frac{2}{3}}{1 – \frac{3}{4} \times \frac{2}{3}}\right)\right]$
$= \tan\left[\tan^{-1}\left(\frac{\frac{9 + 8}{12}}{1 – \frac{1}{2}}\right)\right]$
$= \tan\left[\tan^{-1}\left(\frac{17}{6}\right)\right] = \frac{17}{6}$.

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

(B) $\cos^{-1}\left(\cos\frac{7\pi}{6}\right) \neq \frac{7\pi}{6}$
as the principal value branch of $\cos^{-1}$ is $[0, \pi]$.
So, $\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \cos^{-1}\left[\cos\left(\pi + \frac{\pi}{6}\right)\right]$
$= \cos^{-1}\left(-\cos\frac{\pi}{6}\right) = \cos^{-1}\left[\cos\left(\pi – \frac{\pi}{6}\right)\right]$
$= \frac{5\pi}{6}$.
Hence, $\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$.

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

(D) $\sin\left(\frac{\pi}{3} – \sin^{-1}\left(-\frac{1}{2}\right)\right)$
$= \sin\left(\frac{\pi}{3} + \sin^{-1}\frac{1}{2}\right)$
$= \sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right) = \sin\frac{\pi}{2} = 1$.

21. $\tan^{-1}\sqrt{3} – \cot^{-1}(-\sqrt{3})$ is equal to
Solution:

(B) $\tan^{-1}\sqrt{3} – \cot^{-1}(-\sqrt{3})$
$= \tan^{-1}\sqrt{3} – \left(\pi – \cot^{-1}\sqrt{3}\right)$
$= \frac{\pi}{3} – \pi + \frac{\pi}{6} = -\frac{\pi}{2}$.

Miscellaneous Exercise

Find the value of the following:

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

$\cos^{-1}\left(\cos\frac{13\pi}{6}\right) \neq \frac{13\pi}{6}$
as the range of principal value branch of $\cos^{-1}$ is $[0, \pi]$.
So, $\cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \cos^{-1}\left(\cos\left(2\pi + \frac{\pi}{6}\right)\right)$
$= \cos^{-1}\left(\cos\frac{\pi}{6}\right) = \frac{\pi}{6}$.
$\therefore \cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \frac{\pi}{6}$.

2. $\tan^{-1}\left(\tan\frac{7\pi}{6}\right)$
Solution:

$\tan^{-1}\left(\tan\frac{7\pi}{6}\right) \neq \frac{7\pi}{6}$
as the range of principal value branch of $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
So, $\tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \tan^{-1}\left\{\tan\left(\pi + \frac{\pi}{6}\right)\right\}$
$= \tan^{-1}\left(\tan\frac{\pi}{6}\right) = \frac{\pi}{6}$.
$\therefore \tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \frac{\pi}{6}$.

Prove that:

3. $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$
Solution:

Let $\sin^{-1}\frac{3}{5} = x$.
$\Rightarrow \frac{3}{5} = \sin x \Rightarrow \tan x = \frac{3}{4} \Rightarrow x = \tan^{-1}\frac{3}{4}$.
$\Rightarrow 2\sin^{-1}\frac{3}{5} = 2\tan^{-1}\frac{3}{4}$
$= \tan^{-1}\left(\frac{2 \times \frac{3}{4}}{1 – \left(\frac{3}{4}\right)^2}\right)$
$= \tan^{-1}\left(\frac{\frac{3}{2}}{1 – \frac{9}{16}}\right) = \tan^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)$
$= \tan^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right) = \tan^{-1}\left(\frac{24}{7}\right)$.

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

Let $\sin^{-1}\frac{8}{17} = x \Rightarrow \sin x = \frac{8}{17} \Rightarrow \tan x = \frac{8}{15}$.
Let $\sin^{-1}\frac{3}{5} = y \Rightarrow \frac{3}{5} = \sin y \Rightarrow \tan y = \frac{3}{4} \Rightarrow y = \tan^{-1}\frac{3}{4}$.
$\Rightarrow x + y = \tan^{-1}\frac{8}{15} + \tan^{-1}\frac{3}{4}$
$= \tan^{-1}\left(\frac{\frac{8}{15} + \frac{3}{4}}{1 – \frac{8}{15} \times \frac{3}{4}}\right)$
$= \tan^{-1}\left(\frac{\frac{32 + 45}{60}}{1 – \frac{24}{60}}\right) = \tan^{-1}\left(\frac{77}{36}\right)$.

5. $\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}$
Solution:

Let $x = \cos^{-1}\frac{4}{5}$ and $y = \cos^{-1}\frac{12}{13}$.
$\Rightarrow \cos x = \frac{4}{5}$ and $\cos y = \frac{12}{13}$.
Now, $\sin x = \sqrt{1 – \cos^2 x}$ and $\sin y = \sqrt{1 – \cos^2 y}$.
$\Rightarrow \sin x = \sqrt{1 – \frac{16}{25}} = \frac{3}{5}$
and $\sin y = \sqrt{1 – \frac{144}{169}} = \frac{5}{13}$.
We know that, $\cos(x + y) = \cos x \cos y – \sin x \sin y$
$= \frac{4}{5} \times \frac{12}{13} – \frac{3}{5} \times \frac{5}{13} = \frac{48}{65} – \frac{15}{65} = \frac{33}{65}$.
$\Rightarrow x + y = \cos^{-1}\left(\frac{33}{65}\right)$.
Or, $\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\left(\frac{33}{65}\right)$.

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

Let $x = \cos^{-1}\frac{12}{13}$ and $y = \sin^{-1}\frac{3}{5}$.
Or $\cos x = \frac{12}{13}$ and $\sin y = \frac{3}{5}$.
Now, $\sin x = \sqrt{1 – \cos^2 x} = \sqrt{1 – \frac{144}{169}} = \frac{5}{13}$
and $\cos y = \sqrt{1 – \sin^2 y} = \sqrt{1 – \frac{9}{25}} = \frac{4}{5}$.
We know that, $\sin(x + y) = \sin x \cos y + \cos x \sin y$
$= \frac{5}{13} \times \frac{4}{5} + \frac{12}{13} \times \frac{3}{5} = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}$.
$\Rightarrow x + y = \sin^{-1}\left(\frac{56}{65}\right)$.
Or, $\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\left(\frac{56}{65}\right)$.

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

Let $\sin^{-1}\frac{5}{13} = x$ and $\cos^{-1}\frac{3}{5} = y$.
$\Rightarrow \sin x = \frac{5}{13}$ and $\cos y = \frac{3}{5}$.
Or, $\tan x = \frac{5}{12}$ and $\tan y = \frac{4}{3}$.
$\Rightarrow x + y = \tan^{-1}\frac{5}{12} + \tan^{-1}\frac{4}{3}$
$= \tan^{-1}\left(\frac{\frac{5}{12} + \frac{4}{3}}{1 – \frac{5}{12} \times \frac{4}{3}}\right)$
$= \tan^{-1}\left(\frac{\frac{5 + 16}{12}}{\frac{4}{9}}\right) = \tan^{-1}\left(\frac{21}{12} \times \frac{9}{4}\right)$
$= \tan^{-1}\left(\frac{63}{16}\right)$.

8. $\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}$
Solution:

L.H.S. $= \left(\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{5}\right) + \left(\tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{8}\right)$
$= \tan^{-1}\left(\frac{\frac{1}{3} + \frac{1}{5}}{1 – \frac{1}{3} \times \frac{1}{5}}\right) + \tan^{-1}\left(\frac{\frac{1}{7} + \frac{1}{8}}{1 – \frac{1}{7} \times \frac{1}{8}}\right)$
$= \tan^{-1}\left(\frac{\frac{8}{15}}{\frac{14}{15}}\right) + \tan^{-1}\left(\frac{\frac{15}{56}}{\frac{55}{56}}\right)$
$= \tan^{-1}\frac{8}{14} + \tan^{-1}\frac{15}{55} = \tan^{-1}\frac{4}{7} + \tan^{-1}\frac{3}{11}$
$= \tan^{-1}\left[\frac{\frac{4}{7} + \frac{3}{11}}{1 – \frac{4}{7} \times \frac{3}{11}}\right] = \tan^{-1}\left(\frac{\frac{65}{77}}{\frac{65}{77}}\right)$
$= \tan^{-1}1 = \frac{\pi}{4} =$ R.H.S.

Prove that:

9. $\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1 – x}{1 + x}\right),\;x \in [0, 1]$
Solution:

Putting $x = \tan^2 \theta$, we get
R.H.S. $= \frac{1}{2}\cos^{-1}\left(\frac{1 – x}{1 + x}\right)$
$= \frac{1}{2}\cos^{-1}\left(\frac{1 – \tan^2 \theta}{1 + \tan^2 \theta}\right)$
$= \frac{1}{2}\cos^{-1}(\cos 2\theta) = \frac{1}{2} \times 2\theta = \theta$
$= \tan^{-1}\sqrt{x} =$ L.H.S.
$\therefore \tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1 – x}{1 + x}\right)$.

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

L.H.S. $= \cot^{-1}\left\{\frac{\sqrt{1 + \sin x} + \sqrt{1 – \sin x}}{\sqrt{1 + \sin x} – \sqrt{1 – \sin x}} \times \frac{\sqrt{1 + \sin x} + \sqrt{1 – \sin x}}{\sqrt{1 + \sin x} + \sqrt{1 – \sin x}}\right\}$
$= \cot^{-1}\left\{\frac{(1 + \sin x) + (1 – \sin x) + 2\sqrt{1 – \sin^2 x}}{(1 + \sin x) – (1 – \sin x)}\right\}$
$= \cot^{-1}\left\{\frac{2(1 + \cos x)}{2\sin x}\right\} = \cot^{-1}\left(\frac{1 + \cos x}{\sin x}\right)$
$= \cot^{-1}\left\{\frac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)}\right\}$
$= \cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2} =$ R.H.S.

11. $\tan^{-1}\left(\frac{\sqrt{1 + x} – \sqrt{1 – x}}{\sqrt{1 + x} + \sqrt{1 – x}}\right) = \frac{\pi}{4} – \frac{1}{2}\cos^{-1}x,\;-\frac{1}{\sqrt{2}} \le x \le 1$
Solution:

Putting $x = \cos \theta$, we get
L.H.S. $= \tan^{-1}\left\{\frac{\sqrt{1 + \cos \theta} – \sqrt{1 – \cos \theta}}{\sqrt{1 + \cos \theta} + \sqrt{1 – \cos \theta}}\right\}$
$= \tan^{-1}\left\{\frac{\sqrt{2\cos^2(\theta/2)} – \sqrt{2\sin^2(\theta/2)}}{\sqrt{2\cos^2(\theta/2)} + \sqrt{2\sin^2(\theta/2)}}\right\}$
$= \tan^{-1}\left\{\tan\left(\frac{\pi}{4} – \frac{\theta}{2}\right)\right\}$
$= \frac{\pi}{4} – \frac{\theta}{2} = \frac{\pi}{4} – \frac{1}{2}\cos^{-1}x =$ R.H.S.

12. $\frac{9\pi}{8} – \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3}$
Solution:

$\frac{9\pi}{8} – \frac{9}{4}\sin^{-1}\frac{1}{3}$
$= \frac{9}{4}\left[\frac{\pi}{2} – \sin^{-1}\frac{1}{3}\right] = \frac{9}{4}\cos^{-1}\frac{1}{3}$
$= \frac{9}{4}\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$.
Hence, L.H.S. = R.H.S.

Solve the following equations:

13. $2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)$
Solution:

We have, $2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)$.
$\Rightarrow \tan^{-1}\left[\frac{2\cos x}{1 – \cos^2 x}\right] = \tan^{-1}(2\csc x)$.
$\Rightarrow \tan\left[\tan^{-1}\left(\frac{2\cos x}{\sin^2 x}\right)\right] = 2\csc x$.
$\Rightarrow \frac{2\cos x}{\sin^2 x} = 2\csc x \Rightarrow \cos x = \sin x$.
$\Rightarrow \tan x = 1 \Rightarrow x = \tan^{-1}1 \Rightarrow x = \frac{\pi}{4}$.

14. $\tan^{-1}\left(\frac{1 – x}{1 + x}\right) = \frac{1}{2}\tan^{-1}x,\;(x > 0)$
Solution:

We have, $\tan^{-1}\left(\frac{1 – x}{1 + x}\right) = \frac{1}{2}\tan^{-1}x,\;(x > 0)$.
$\Rightarrow \tan^{-1}1 – \tan^{-1}x = \frac{1}{2}\tan^{-1}x$.
$\Rightarrow \frac{3}{2}\tan^{-1}x = \tan^{-1}1 = \frac{\pi}{4}$.
$\Rightarrow \tan^{-1}x = \frac{\pi}{4} \times \frac{2}{3} = \frac{\pi}{6}$.
$\Rightarrow x = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

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

(D) Let $\tan^{-1}x = \theta \Rightarrow x = \tan \theta,\;\text{where}\;\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$\therefore \sin(\tan^{-1}x) = \sin \theta$.
Now, $\sin \theta = \frac{1}{\csc \theta} = \frac{1}{\sqrt{1 + \cot^2 \theta}} = \frac{1}{\sqrt{1 + \frac{1}{\tan^2 \theta}}}$
$= \frac{x}{\sqrt{x^2 + 1}}$.

16. If $\sin^{-1}(1 – x) – 2\sin^{-1}x = \frac{\pi}{2}$, then $x$ is equal to
Solution:

(C) $\sin^{-1}(1 – x) – 2\sin^{-1}x = \frac{\pi}{2}$.
$\Rightarrow \sin^{-1}(1 – x) = \frac{\pi}{2} + 2\sin^{-1}x$.
$\Rightarrow (1 – x) = \sin\left(\frac{\pi}{2} + 2\sin^{-1}x\right) = \cos(2\sin^{-1}x)$.
$\Rightarrow 1 – x = \cos(\cos^{-1}(1 – 2x^2))$.
$\Rightarrow 1 – x = 1 – 2x^2 \Rightarrow 2x^2 – x = 0 \Rightarrow x = 0, \frac{1}{2}$.
But, $x = \frac{1}{2}$ does not satisfy the equation. So, $x = 0$.

17. $\tan^{-1}\left(\frac{x}{y}\right) – \tan^{-1}\left(\frac{x – y}{x + y}\right)$ is equal to
Solution:

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

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