Question

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

• A. ${\tan }^2(\alpha /2)$
• B. ${\cot }^2(\alpha /2)$
• C. $\tan \alpha$
• D. $\cot (\alpha /2)$

Answer 1

Answer: (A)

We have,
${\cot }^{-1}(\sqrt{\cos \alpha })-{\tan }^{-1}(\sqrt{\cos \alpha })=x$
$\frac{\pi }{2}-{\tan }^{-1}\sqrt{\cos \alpha }-{\tan }^{-1}\sqrt{\cos \alpha }=x$
$\Rightarrow 2{\tan }^{-1}(\sqrt{\cos \alpha })=\frac{\pi }{2}-x$
$\Rightarrow {\cos }^{-1}\left(\frac{1-\cos \alpha }{1+\cos \alpha }\right)=\frac{\pi }{2}-x$
$\left[\because 2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$
$\frac{2{\sin }^2\frac{\alpha }{2}}{2{\cos }^2\frac{\alpha }{2}}=\cos \left(\frac{\pi }{2}-x\right)$
$\Rightarrow {\tan }^2\frac{\alpha }{2}=\sin x$
$\therefore \sin x={\tan }^2\frac{\alpha }{2}$
Note There is a correction in question. It is $\sin x$ not