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] \[{{\cot }^{-1}}(\sqrt{\cos \alpha })-ta{{n}^{-1}}(\sqrt{\cos \alpha })=x\] Or \[{{\tan }^{-1}}\left( \frac{1}{\sqrt{\cos \alpha }} \right)-{{\tan }^{-1}}(\sqrt{\cos \alpha })=x\] Or \[{{\tan }^{-1}}\frac{\frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}{1+\frac{1}{\sqrt{\cos \alpha }}\sqrt{\cos \alpha }}=x\] Or \[{{\tan }^{-1}}\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}=x\] Or \[\tan x=\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}=x\] Or \[\cot x=\frac{2\sqrt{\cos \alpha }}{1-\cos \alpha }\] Or \[\text{cosec}\,x=\sqrt{1+\frac{4\cos \alpha }{{{(1-cos\alpha )}^{2}}}}=\frac{1+\cos \alpha }{1-\cos \alpha }\] Or \[\sin x=\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{2{{\sin }^{2}}(\alpha /2)}{2{{\cos }^{2}}(\alpha /2)}={{\tan }^{2}}\frac{\alpha }{2}\]
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