Question

The value of $\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]$ is

• A. $\frac{1}{\sqrt{x^2+2}}$
• B. $\sqrt{\frac{x^2+2}{x^2+1}}$
• C. $\sqrt{\frac{x^2+1}{x^2+2}}$
• D. $\frac{1}{\sqrt{x^2+1}}$

Answer 1

Answer: (C)

We have, $\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]$
Let $\cot ^{-1} x=\alpha$
$\Rightarrow \cot \alpha=x$
$\Rightarrow \operatorname{cosec} \alpha=\sqrt{1+x^2}$
$\Rightarrow \sin \alpha=\frac{1}{\sqrt{1+x^2}}$
$\Rightarrow \alpha=\sin ^{-1} \frac{1}{\sqrt{1+x^2}}$
Hence, $\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]$
$=\cos \left[\tan ^{-1}\left\{\sin \left(\sin ^{-1} \frac{1}{\sqrt{1+x^2}}\right)\right\}\right]$
$=\cos \left[\tan ^{-1} \frac{1}{\sqrt{1+x^2}}\right]$
$ = \cos\left[\sqrt{\frac{x^2 + 1}{x^2+2}}\right]$
$[\because$ If we let $\tan^{-1} \frac{1}{\sqrt{1+x^2}} = \beta$
$\tan \beta = \frac{1}{\sqrt{1+ x^2}}$
$\sec \beta = \sqrt{1 + \frac{1}{1+x^2}} = \sqrt{\frac{x^2+2}{x^2 + 1}}$
$\cos \beta = \sqrt{\frac{x^ + 1}{x^2 + 2}}]$
$=\sqrt{\frac{x^2+1}{x^2+2}}$