Question

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

Answer 1

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

Let $x=\cot \theta$

L.H.S$=\cos \left[{\tan }^{-1}\left\{\sin \theta \right\}\right]$

$=\cos \left[{\tan }^{-1}\left\{\frac{1}{\sqrt{1+{\cot }^2\theta }}\right\}\right]$

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

Let ${\theta }_1={\tan }^{-1}\left\{\frac{1}{\sqrt{1+x^2}}\right\}$

$\Rightarrow \tan {\theta }_1=\frac{1}{\sqrt{1+x^2}}$

$\Rightarrow \cos {\theta }_1=\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}$

$\Rightarrow {\theta }_1={\cos }^{-1}\left(\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}\right)$

Now put ${\theta }_1$ in equation $\left(1\right)$ we get

$\cos \left[{\cos }^{-1}\left(\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}\right)\right]=\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}=$R.H.S

Hence proved.