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 \[\Delta ABC,\] \[A{{C}^{2}}=A{{B}^{2}}+C{{B}^{2}}\] \[={{(2\sqrt{\cos \alpha })}^{2}}+{{(1-\cos \alpha )}^{2}}\] \[=4\cos \alpha +1+{{\cos }^{2}}\alpha -2\cos \alpha \] \[\Rightarrow \]\[AC=1+\cos \alpha \] \[\therefore \]\[\sin x=\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{2{{\sin }^{2}}\frac{\alpha }{2}}{2{{\cos }^{2}}\frac{\alpha }{2}}\] \[\therefore \]\[\sin x={{\tan }^{2}}\frac{\alpha }{2}\]You need to login to perform this action.
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