Question

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

• A. ${\tan }^2\left(\frac{\alpha }{2}\right)$
• B. $co{t}^2\left(\frac{\alpha }{2}\right)$
• C. $\tan \alpha$
• D. $\cot \left(\frac{\alpha }{2}\right)$

Answer 1

Answer: (A)

${\cot }^{-1}(\sqrt{\cos \alpha })-{\tan }^{-1}(\sqrt{\cos \alpha })=x$ ${\tan }^{-1}\left(\frac{1}{\sqrt{\cos }\alpha }\right)-{\tan }^{-1}(\sqrt{\cos \alpha })=x$ $\Rightarrow$ ${\tan }^{-1}\frac{\frac{1}{\sqrt{\cos }\alpha }-\sqrt{\cos \alpha }}{1+\frac{1}{\sqrt{\cos \alpha }}.\sqrt{\cos \alpha }}=x$ $\Rightarrow$ ${\tan }^{-1}\frac{1-\cos \alpha }{2\sqrt{\cos }\alpha }=x$ $\Rightarrow$ $\tan x=\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}$ $\sin x=\frac{1-\cos \alpha }{1+\cos \alpha }$ $=\frac{1-(1-2{\sin }^2\alpha /2)}{1+2{\cos }^2\alpha /2-1}$ $\sin x={\tan }^2\frac{\alpha }{2}$