$${\cot ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x,$$ then $$sinx = $$
A
$${\tan ^2}\left( {{\alpha \over 2}} \right)$$
B
$${\cot ^2}\left( {{\alpha \over 2}} \right)$$
D
$$cot\left( {{\alpha \over 2}} \right)$$
Explanation
Given that,
$${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) - {\tan ^{ - 1}}\left( {\cos x} \right) = x\,\,\,\,...\left( 1 \right)$$
We know,
$${\cot ^{ - 1}}x + {\tan ^{ - 1}}x = {\pi \over 2}$$
$$\therefore$$ $$\,\,\,$$ $${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) + {\tan ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {\pi \over 2}\,\,\,...\left( 2 \right)$$
Adding $$(1)$$ and $$(2),$$ we get,
$$2{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = x + {\pi \over 2}$$
$$ \Rightarrow \,\,{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {x \over 2} + {\pi \over 4}$$
$$ \Rightarrow \,\,\sqrt {\cos x} = \cot \left( {{x \over 2} + {\pi \over 4}} \right)$$
$$ \Rightarrow \,\,\sqrt {\cos x} = {{\cot {x \over 2} - 1} \over {1 + \cot {x \over 2}}}$$
$$ \Rightarrow \,\,\sqrt {\cos x} = {{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}}$$
Squaring both sides we get,
$$ \Rightarrow \,\,\cos x = {{1 - 2\sin {x \over 2}\cos {x \over 2}} \over {1 + 2\sin {x \over 2}\cos {x \over 2}}}$$
$$ \Rightarrow \,\,\cos x = {{1 - \sin x} \over {1 + \sin x}}$$
$$ \Rightarrow \,\,{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}} = {{1 - \sin x} \over {1 + \sin x}}$$
Applying compounds and dividendo rule,
$$ \Rightarrow \,\,{{2\sin x} \over 2} = {{2{{\tan }^2}{x \over 2}} \over 2}$$
$$ \Rightarrow \,\,\sin x = {\tan ^2}{x \over 2}$$