2016/013) Prove that $\cos(\tan^{-1}(\sin(\cot^{-1} x)))= \sqrt{\frac{x^2+1}{x^2+2}}$

Let $\cot^{-1} x = y$
so $\cot y = x$
so $\csc^2 y = x^2+1$
so $\sin  y = \frac{1}{\sqrt{x^2+1}}$
so $\sin\cot^{-1} x = \frac{1}{\sqrt{x^2+1}}$
similarly  $\cos \tan ^{-1}x = \sqrt{\frac{1}{x^2+1}}$
Hence $\cos(\tan^{-1}(\sin(\cot^{-1} x)))$
$=\cos(\tan^{-1}\frac{1}{\sqrt{x^2+1}})$
$=\frac{1}{1+\frac{1}{x^2+1}}$
$=\frac{1}{\sqrt{1+\frac{1}{x^2+1}}}$
$=\frac{1}{\sqrt{\frac{x^2+2}{x^2+1}}}$
$=\sqrt{\frac{x^2+1}{x^2+2}}$