Question

Match the inverse functions in column I with their principal values in column II and choose the correct option from the codes given below.

Column I		Column II
A	$ \cot ^{-1} \sqrt{3}$	1	$\frac{2 \pi}{3}$
B	$\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$	2	$-\frac{\pi}{4}$
C	$\operatorname{cosec}^{-1}(-\sqrt{2})$	3	$\frac{3 \pi}{4}$
D	$\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$	4	$\frac{\pi}{6}$

• A. A - (4), B - (3),C - (2), D - (1)
• B. A - (3), B - (4),C - (2), D - (1)
• C. A - (2), B - (3),C - (4), D - (1)
• D. A - (4), B - (3),C - (1), D - (2)

Answer 1

Answer: (A)

A. Let $\cot ^{-1}(\sqrt{3})=\theta \Rightarrow \cot \theta=\sqrt{3}$
We know that, the range of principal value branch of $\cot ^{-1} \theta$ is $(0, \pi)$.
$\because \cot \theta=\sqrt{3}=\cot \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6} $ where, $\theta \in(0, \pi)$
$\Rightarrow \cot ^{-1}(\sqrt{3})=\frac{\pi}{6}$
Hence, the principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$.
B. Let $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\theta \Rightarrow \cos \theta=-\frac{1}{\sqrt{2}}$
We know that the range of principal value branch of $\cos ^{-1} \theta$ is $[0, \pi]$
$\because \cos \theta=-\frac{1}{\sqrt{2}} =-\cos \frac{\pi}{4}=\cos \left(\pi-\frac{\pi}{4}\right) $
$\Rightarrow {[\because \cos (\pi-\theta)=-\cos \theta)] }$
$\theta =\frac{3 \pi}{4} \text { where, } \theta \in[0, \pi] $
$\Rightarrow \cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right) =\frac{3 \pi}{4}$
Hence, the principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ is $\frac{3 \pi}{4}$.
C. Let $\operatorname{cosec}^{-1}(-\sqrt{2})=\theta \Rightarrow \operatorname{cosec} \theta=-\sqrt{2}$
We know that, the range of principal value branch of $\operatorname{cosec}^{-1} \theta$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$.
$\because \operatorname{cosec} \theta=-\sqrt{2}=-\operatorname{cosec} \frac{\pi}{4}=\operatorname{cosec}\left(-\frac{\pi}{4}\right) $
$[\because \operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta] $
$ \Rightarrow \theta=-\frac{\pi}{4} \text { where, } \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} $
$ \Rightarrow \operatorname{cosec}^{-1}(-\sqrt{2})=-\frac{\pi}{4}$
Hence, the principal value of $\operatorname{cosec}^{-1}(-\sqrt{2})$ is $-\frac{\pi}{4}$.
D. Let $\cos ^{-1}\left(\frac{1}{2}\right)=x \Rightarrow \cos x=\frac{1}{2}=\cos \frac{\pi}{3}$
$\Rightarrow x=\frac{\pi}{3} \in[0, \pi] $ (principal value branch)
Again, let $\sin ^{-1}\left(\frac{1}{2}\right)=y$
$ \Rightarrow \sin y=\frac{1}{2}=\sin \frac{\pi}{6} $
$ \Rightarrow y=\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $ (principal value branch)
$ \therefore \cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)=x+2 y $
$ =\frac{\pi}{3}+2 \times \frac{\pi}{6}=\frac{2 \pi}{3} $