Question

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

• 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

The correct option is A $\sqrt{\frac{1+x^2}{2+x^2}}$

Let ${\tan }^{-1}x=\alpha \Rightarrow \tan \alpha =x$

So $\cos \alpha =\frac{1}{\sqrt{1+x^2}}$

$\therefore \sin \left[{\cot }^{-1}\left\{\cos \left({\tan }^{-1}x\right)\right\}\right]=\sin \left({\cot }^{-1}\frac{1}{\sqrt{1+x^2}}\right)$

Let ${\cot }^{-1}\frac{1}{\sqrt{1+x^2}}=\beta \Rightarrow \cot \beta =\frac{1}{\sqrt{1+x^2}}$

$\therefore \sin \beta =\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}$

$\therefore \sin \beta =\sin \left[{\cot }^{-1}\left\{\cos \left({\tan }^{-1}x\right)\right\}\right]=\sqrt{\frac{1+x^2}{2+x^2}}$