\(\text {L.H.S. }=\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]\)
माना \(\cot ^{-1} x=\theta \Rightarrow x=\cot \theta\)
\(\therefore \operatorname{cosec} \theta=\sqrt{1+\cot ^2 \theta}=\sqrt{1+x^2} \)
\(\therefore \sin \theta=\frac{1}{\sqrt{1+x^2}}\)
अब \(\text{L.H.S.} =\cos \left[\tan ^{-1}(\sin \theta)\right]=\cos \left[\tan ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right]\)
फिर माना कि \(\tan ^{-1} \frac{1}{\sqrt{1+x^2}}=\alpha \Rightarrow \tan \alpha=\frac{1}{\sqrt{1+x^2}}\)
\(\therefore \sec ^2 \alpha=1+\tan ^2 \alpha=1+\frac{1}{1+x^2}=\frac{x^2+2}{x^2+1} \)
\(\Rightarrow \sec \alpha=\sqrt{\frac{x^2+2}{x^2+1}} \Rightarrow \cos \alpha=\sqrt{\frac{x^2+1}{x^2+2}}\)
अब \(\text{L.H.S.} =\cos \alpha=\sqrt{\frac{x^2+1}{x^2+2}}\)
