Question

$\sin [{\cot }^{-1}\{\cos ({\tan }^{-1}x)\}]=1$

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

Answer 1

Answer: (A)

$cos\left(ta{n}^{-1}x\right)=cos\theta$
where $\theta =ta{n}^{-1}x$ i.e., $tan\theta =x$

$\therefore cos\theta =\frac{1}{\sqrt{1+x^2}}$
$\therefore cos\left(ta{n}^{-1}x\right)=\frac{1}{\sqrt{1+x^2}}$
$co{t}^{-1}\left[cos\left(ta{n}^{-1}x\right)\right]=co{t}^{-1}\left[\frac{1}{\sqrt{1+x^2}}\right]=t$
$\Rightarrow cott=\frac{1}{\sqrt{1+x^2}}$

$\therefore sin\left[co{t}^{-1}\left\{cos\left(ta{n}^{-1}x\right)\right\}\right]=\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}$