question_answer

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

A)  \[{{\cot }^{2}}\left( \frac{\alpha }{2} \right)\]                    

B)  \[{{\tan }^{2}}\left( \frac{\alpha }{2} \right)\]

C)  \[\tan \alpha \]

D)  \[{{\cot }^{2}}\left( \frac{\alpha }{2} \right)\]

Correct Answer: B

Solution :

Given that,    \[{{\cot }^{-1}}(\sqrt{\cos \alpha })-{{\tan }^{-1}}(\sqrt{\cos \alpha })=x\] \[\therefore \]\[{{\tan }^{-1}}\left( \frac{1}{\sqrt{\cos \alpha }} \right)-{{\tan }^{-1}}(\sqrt{\cos \alpha })=x\] \[\Rightarrow \]\[{{\tan }^{-1}}\left( \frac{\frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}{1+\frac{1}{\sqrt{\cos \alpha }}.\sqrt{\cos \alpha }} \right)=x\] \[\Rightarrow \]\[{{\tan }^{-1}}\left( \frac{1-\cos \alpha }{2\sqrt{\cos \alpha }} \right)=x\] \[\Rightarrow \]\[\tan x=\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}\] In \[\Delta ABC,\] \[A{{C}^{2}}=A{{B}^{2}}+C{{B}^{2}}\] \[={{(2\sqrt{\cos \alpha })}^{2}}+{{(1-\cos \alpha )}^{2}}\] \[=4\cos \alpha +1+{{\cos }^{2}}\alpha -2\cos \alpha \] \[\Rightarrow \]\[AC=1+\cos \alpha \] \[\therefore \]\[\sin x=\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{2{{\sin }^{2}}\frac{\alpha }{2}}{2{{\cos }^{2}}\frac{\alpha }{2}}\] \[\therefore \]\[\sin x={{\tan }^{2}}\frac{\alpha }{2}\]