Chapter 2: Inverse Trigonometric Functions

Ans: Let’s consider ${\tan ^{ - 1}}(1) = x$. Then, $\tan x = 1 = \tan \left( {\dfrac{\pi }{4}} \right)$.

$\therefore {\tan ^{ - 1}}(1) = \dfrac{\pi }{4}$

Let’s assume,${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y$.

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

$\therefore {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{{2\pi }}{3}$

Let’s again assume that ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = z$.

Then, $\sin z =  - \dfrac{1}{2} =  - \sin \left( {\dfrac{\pi }{6}} \right) = \sin \left( { - \dfrac{\pi }{6}} \right)$.

$\therefore {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) =  - \dfrac{\pi }{6}$

$\therefore {\tan ^{ - 1}}(1) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$

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

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

Ans: Let’s consider,${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = x$.

Then, $\cos x = \dfrac{1}{2} = \cos \left( {\dfrac{\pi }{3}} \right)$.

$\therefore {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3}$

Let’s assume ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = y.$

Then, $\sin y = \dfrac{1}{2} = \sin \left( {\dfrac{\pi }{6}} \right)$.

$\therefore {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}$

$\therefore {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3} + 2 \times \dfrac{\pi }{6} = \dfrac{\pi }{3} + \dfrac{\pi }{3} = \dfrac{{2\pi }}{3}$

Ans: It is given that ${\sin ^{ - 1}}x = y$. As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$. 

Therefore, $ - \dfrac{\pi }{2} \leqslant y \leqslant \dfrac{\pi }{2}$.

Hence option (B) is correct.

Ans: Let’s consider, ${\tan ^{ - 1}}\sqrt 3  = x.$.

Then, $\tan x = \sqrt 3  = \tan \dfrac{\pi }{3}$

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$. $\therefore {\tan ^{ - 1}}\sqrt 3  = \dfrac{\pi }{3}$

Let assume, ${\sec ^{ - 1}}( - 2) = y$.

Then, $\sec y =  - 2 =  - \sec \left( {\dfrac{\pi }{3}} \right) = \sec \left( {\pi  - \dfrac{\pi }{3}} \right) = \sec \dfrac{{2\pi }}{3}$.

As we know that the range of the principal value branch of sec $^1$ is $[0,\pi ] - \left\{ {\dfrac{\pi }{2}} \right\}$. $\therefore {\sec ^{ - 1}}( - 2) = \dfrac{{2\pi }}{3}$

Thus, ${\tan ^{ - 1}}(\sqrt 3 ) - {\sec ^{ - 1}}( - 2)$

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