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 ${\cot }^{-1}x=\alpha$
$\Rightarrow \cot \alpha =x$
$\Rightarrow \mathrm{cosec}\alpha =\sqrt{1+x^2}$
$\Rightarrow \sin \alpha =\frac{1}{\sqrt{1+x^2}}$
$\Rightarrow \alpha ={\sin }^{-1}\frac{1}{\sqrt{1+x^2}}$
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 ${\tan }^{-1}\frac{1}{\sqrt{1+x^2}}=\beta$
$\tan \beta =\frac{1}{\sqrt{1+x^2}}$
$\sec \beta =\sqrt{1+\frac{1}{1+x^2}}=\sqrt{\frac{x^2+2}{x^2+1}}$
$\cos \beta =\sqrt{\frac{x^{+}1}{x^2+2}}]$
$=\sqrt{\frac{x^2+1}{x^2+2}}$