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