question_answer

If \[y={{\cot }^{-1}}(\sqrt{\cos \,x})-ta{{n}^{-1}}(\sqrt{\cos \,x}),\] prove that \[\sin y={{\tan }^{2}}\frac{x}{2}.\]

Answer:

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