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 :
\[{{\tan }^{-1}}\left[ \frac{1}{\sqrt{\cos \alpha }} \right]-{{\tan }^{-1}}\left[ \sqrt{\cos \alpha } \right]=x\] Þ\[{{\tan }^{-1}}\left[ \frac{\frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}{1+\frac{\sqrt{\cos \alpha }}{\sqrt{\cos \alpha }}} \right]=x\]Þ \[\tan x=\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}\] \[\therefore \sin x=\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{2{{\sin }^{2}}\frac{\alpha }{2}}{2{{\cos }^{2}}\frac{\alpha }{2}}={{\tan }^{2}}\left( \frac{\alpha }{2} \right)\].
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