A) \[{{\tan }^{2}}\left( \frac{\alpha }{2} \right)\]
B) \[{{\cot }^{2}}\left( \frac{\alpha }{2} \right)\]
C) \[\tan \alpha \]
D) \[\cot \left( \frac{\alpha }{2} \right)\]
Correct Answer: A
Solution :
\[{{\cot }^{-1}}(\sqrt{\cos \alpha })-{{\tan }^{-1}}(\sqrt{\cos \alpha })=x\] \[{{\tan }^{-1}}\left( \frac{1}{\sqrt{\cos \alpha }} \right)-{{\tan }^{-1}}(\sqrt{\cos \alpha })=x\] \[\Rightarrow \] \[{{\tan }^{-1}}\frac{\frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}{1+\frac{1}{\sqrt{\cos \alpha }}\sqrt{\cos \alpha }}\] \[\Rightarrow \] \[{{\tan }^{-1}}\frac{1-\cos \alpha }{2\sqrt{\cos }\alpha }=x\] \[\Rightarrow \] \[\tan x=\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}\] \[\Rightarrow \] \[\cot x=\frac{2\sqrt{\cos \alpha }}{1-\cos \alpha }\] \[\Rightarrow \] \[\cos \text{ec}x=\frac{1+\cos \alpha }{1-\cos \alpha }\] \[\therefore \] \[\sin x=\frac{1-\cos \alpha }{1+\cos \alpha }\] \[=\frac{1-\left( 1-2{{\sin }^{2}}\frac{\alpha }{2} \right)}{1+2{{\cos }^{2}}\frac{\alpha }{2}-1}\] \[\Rightarrow \] \[\sin x={{\tan }^{2}}\frac{\alpha }{2}\]
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