Answer:
We have, \[y={{\cot }^{-1}}(\sqrt{\cos x})-ta{{n}^{-1}}(\sqrt{\cos x})\] \[\Rightarrow \] \[y=\frac{\pi }{2}-{{\tan }^{-1}}(\sqrt{\cos x})-ta{{n}^{-1}}(\sqrt{\cos x})\] \[\Rightarrow \] \[y=\frac{\pi }{2}-2{{\tan }^{-1}}(\sqrt{\cos x})\] \[\Rightarrow \] \[y=\frac{\pi }{2}-{{\cos }^{-1}}\left\{ \frac{1-{{(\sqrt{\cos x})}^{2}}}{1+{{(\sqrt{\cos x})}^{2}}} \right\}\] \[\left[ \because 2{{\tan }^{-1}}x={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right]\] \[\Rightarrow \] \[y=\frac{\pi }{2}-{{\cos }^{-1}}\left( \frac{1-\cos \,x}{1+\cos \,x} \right)\] \[\Rightarrow \] \[y=\frac{\pi }{2}-{{\cos }^{-1}}\left( {{\tan }^{2}}\frac{x}{2} \right)\] \[\Rightarrow \] \[{{\cos }^{-1}}\left( {{\tan }^{2}}\frac{x}{2} \right)=\frac{\pi }{2}-y\] \[\Rightarrow \] \[{{\tan }^{2}}\frac{x}{2}=\cos \left( \frac{\pi }{2}-y \right)\] \[\Rightarrow \] \[{{\tan }^{2}}\frac{x}{2}=\sin \,y\] Hence proved.
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