Question
Prove the following:
$\cos [\tan^{–1} {\sin (\cot^{–1} x)}]$ = $\sqrt{\frac{\text{1 + x}^{2}}{\text{2 + x}^{2}}}$.

Answer

$LHS = \cos [\tan^{-1} {\sin (\cot^{–1} x)}]$
= $\cos\Bigg[\tan^{-1}\left\{\sin\Bigg(\sin^{-1}\frac{1}{\sqrt{\text{1 + x}^{2}}}\Bigg)\right\}\Bigg]$
= cos $\Bigg[\tan^{-1}\Bigg(\frac{1}{\sqrt{\text{1 + x}^{2}}}\Bigg)\Bigg]=\cos\Bigg[\cos^{-1}\frac{\sqrt{\text{1 + x}^{2}}}{\sqrt{\text{2 + x}^{2}}}\Bigg]$
$=\frac{\sqrt{\text{1 + x}^{2}}}{\sqrt{\text{2 + x}^{2}}}=\text{R.H.S.}$

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