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

Answer 1

Answer: (A)

We know that
$\cot ^{-1}(\sqrt{\cos \alpha})+\tan ^{-1}(\sqrt{\cos \alpha})=\frac{\pi}{2} \ldots \text { (i) }$
and given that
$\cot ^{-1} \sqrt{\cos \alpha-\tan ^{-1} \sqrt{\cos \alpha}}=x \ldots \text { (ii) }$
On adding Eqs. (i) and (ii), we get
$2 \cot ^{-1}(\sqrt{\cos \alpha}) =\frac{\pi}{2}+x $
$\Rightarrow \sqrt{\cos \alpha} =\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)$
$\Rightarrow \sqrt{\cos \alpha}=\frac{\cot \frac{x}{2}-1}{1+\cot \frac{x}{2}}$
$\Rightarrow \cos \alpha=\frac{1-\sin x}{1+\sin x}$
$\Rightarrow \frac{1-\tan ^{2} \frac{\alpha}{2}}{1+\tan ^{2} \frac{\alpha}{2}}=\frac{1-\sin x}{1+\sin x}$
$\Rightarrow \sin x=\tan ^{2} \frac{\alpha}{2}$
Alternate solution
$\because \cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$
$\Rightarrow \tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x$
$\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos (x}} \cdot \sqrt{\cos \alpha}}\right)=x$
$\Rightarrow \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=\tan x$
$\Rightarrow \cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha}$
$cosec x=\sqrt{1+\cot ^{2} x}$
$\therefore cosec x=\frac{1+\cos \alpha}{1-\cos \alpha}$
$\Rightarrow \sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\tan ^{2} \frac{\alpha}{2}$