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