Given that, ${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) - {\tan ^{ - 1}}\left( \sqrt{\cos x} \right) = x\,\,\,\,...\left( 1 \right)$ 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, $\Rightarrow \,\,{{2\sin x} \over 2} = {{2{{\tan }^2}{x \over 2}} \over 2}$ $\Rightarrow \,\,\sin x = {\tan ^2}{x \over 2}$ Other Method : $\begin{aligned} & \text { Since, } \cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}\right)=x \\\\ & \Rightarrow \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=\tan x \\\\ & \Rightarrow \cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha} \\\\ & \because \operatorname{cosec} x=\sqrt{1+\cot { }^2 x} \\\\ & \therefore \operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha} \\\\ & \Rightarrow \sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\tan ^2 \frac{\alpha}{2} \end{aligned}$