\[{{\cot }^{-1}}(\sqrt{\cos \alpha })-ta{{n}^{-1}}(\sqrt{\cos \alpha })=x\], then sin x is equal to

\[{{\cot }^{2}}\frac{\alpha }{2}\]

\[{{\tan }^{2}}\frac{\alpha }{2}\]

\[{{\cot }^{-1}}(\sqrt{\cos \alpha })-ta{{n}^{-1}}(\sqrt{\cos \alpha }

1.	\[{{\cot }^{-1}}(\sqrt{\cos \alpha })-ta{{n}^{-1}}(\sqrt{\cos \alpha })=x\], then sin x is equal to
A.	\[{{\tan }^{2}}\frac{\alpha }{2}\]
B.	\[{{\cot }^{2}}\frac{\alpha }{2}\]
C.	\[\tan \alpha \]
D.	\[\cot \frac{\alpha }{2}\]
Answer» B. \[{{\cot }^{2}}\frac{\alpha }{2}\]

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\[{{\cot }^{-1}}(\sqrt{\cos \alpha })-ta{{n}^{-1}}(\sqrt{\cos \alpha })=x\], then sin x is equal to

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