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

Let ${\cot }^{-1}\mathrm{x}=\mathrm{\theta }\Rightarrow \cot \mathrm{\theta }=\mathrm{x}$

$\therefore \sin \left({\cot }^{-1}\mathrm{x}\right)=\sin \mathrm{\theta }=\frac{1}{\mathrm{cosec}\mathrm{\theta }}=\frac{1}{\sqrt{1+{\cot }^2\mathrm{\theta }}}=\frac{1}{\sqrt{1+{\mathrm{x}}^2}}$

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

$=\cos \left[{\tan }^{-1}\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\right]=\cos \mathrm{\phi }$

$\left(\text{ Put }{\tan }^{-1}\left\{\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\right\}=\mathrm{\varphi }\Rightarrow \tan \mathrm{\varphi }=\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\right)$

$=\frac{1}{\sec \mathrm{\varphi }}=\frac{1}{\sqrt{1+{\tan }^2\mathrm{\varphi }}}=\frac{1}{\sqrt{1+\frac{1}{1+{\mathrm{x}}^2}}}=\sqrt{\frac{1+{\mathrm{x}}^2}{2+{\mathrm{x}}^2}}$