Question

If $x\neq n \pi ,\, x \neq\,(2n+1)\frac {\pi}{2}.n\in Z, $ then $\frac {Sin^{-1}(Cos x) + Cos^{-1}(Sin x)}{Tan ^{-1}(Cot x)+ Cot^{-1}(Tan x)} $ =

• A. $\frac {\pi} {6}$
• B. $\frac {\pi} {3}$
• C. $\frac {\pi} {2}$
• D. $\frac {\pi} {4}$

Answer 1

The Correct Option is D

$x \neq n \pi, x \neq(2 n+1) \frac{\pi}{2}, n \in Z$
Then, $\frac{\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)}{\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x)}$
$=\frac{\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-x\right)\right\}+\cos ^{-1}\left\{\cos \left(\frac{\pi}{2}-x\right)\right\}}{\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-x\right)\right\}+\cot ^{-1}\left\{\cot \left(\frac{\pi}{2}-x\right)\right\}}$
$=\frac{\left(\frac{\pi}{2}-x\right)+\left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)+\left(\frac{\pi}{2}-x\right)}=\left(\frac{\pi-2 x}{\pi-2 x}\right)=1$
But from the option we take $x=\frac{\pi}{4}$
$=\frac{\sin ^{-1}\left(\cos \frac{\pi}{4}\right)+\cos ^{-1}\left(\sin \frac{\pi}{4}\right)}{\tan ^{-1}\left(\cot \frac{\pi}{4}\right)+\cot ^{-1}\left(\tan \frac{\pi}{4}\right)}$
$=\frac{\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)}{\tan ^{-1}(1)+\cot ^{-1}(1)}$
$\begin{Bmatrix}\sin^{-1}x +\cos^{-1}&x=\frac{\pi}{2}\\ \because&\\ \tan^{-1} x +cot^{-1}&x=\frac{\pi}{2}\end{Bmatrix}$
$=\frac{\pi / 2}{\pi / 2}=1$