Question
$${\cot ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x,$$ then $$\sin x = $$
A.
$${\tan ^2}\left( {\frac{\alpha }{2}} \right)$$
B.
$${\cot ^2}\left( {\frac{\alpha }{2}} \right)$$
C.
$$\tan \alpha $$
D.
$${\cot}\left( {\frac{\alpha }{2}} \right)$$
Answer :
$${\tan ^2}\left( {\frac{\alpha }{2}} \right)$$
Solution :
$$\eqalign{
& {\cot ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x \cr
& {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt {\cos \alpha } }}} \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x \cr
& \Rightarrow \,\,{\tan ^{ - 1}}\frac{{\frac{1}{{\sqrt {\cos \alpha } }} - \sqrt {\cos \alpha } }}{{1 + \frac{1}{{\sqrt {\cos \alpha } }}.\sqrt {\cos \alpha } }} = x \cr
& \Rightarrow \,\,{\tan ^{ - 1}}\frac{{1 - \cos \alpha }}{{2\sqrt {\cos \alpha } }} = x \cr
& \Rightarrow \,\,\tan x = \frac{{1 - \cos \alpha }}{{2\sqrt {\cos \alpha } }}\,\,{\text{or }}\cot x = \frac{{2\sqrt {\cos \alpha } }}{{1 - \cos \alpha }} \cr
& \Rightarrow \,\,\sin x = \frac{{1 - \cos \alpha }}{{1 + \cos \alpha }} \cr
& = \frac{{1 - \left( {1 - 2{{\sin }^2}\frac{\alpha }{2}} \right)}}{{1 + 2{{\cos }^2}\frac{\alpha }{2} - 1}}\,\,{\text{or }}\sin x = {\tan ^2}\frac{\alpha }{2} \cr} $$