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
${\cot }^{-1}(\sqrt{\cos \alpha })+{\tan }^{-1}(\sqrt{\cos \alpha })=\frac{\pi }{2}\dots \text{ (i) }$
and given that
${\cot }^{-1}\sqrt{\cos \alpha -{\tan }^{-1}\sqrt{\cos \alpha }}=x\dots \text{ (ii) }$
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
$\because {\cot }^{-1}(\sqrt{\cos \alpha })-{\tan }^{-1}(\sqrt{\cos \alpha })=x$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{\sqrt{\cos \alpha }}\right)-{\tan }^{-1}(\sqrt{\cos \alpha })=x$
$\Rightarrow {\tan }^{-1}\left(\frac{\frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}{1+\frac{1}{\sqrt{\cos (x}}\cdot \sqrt{\cos \alpha }}\right)=x$
$\Rightarrow \frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}=\tan x$
$\Rightarrow \cot x=\frac{2\sqrt{\cos \alpha }}{1-\cos \alpha }$
$cosecx=\sqrt{1+{\cot }^{2}x}$
$\therefore cosecx=\frac{1+\cos \alpha }{1-\cos \alpha }$
$\Rightarrow \sin x=\frac{1-\cos \alpha }{1+\cos \alpha }={\tan }^2\frac{\alpha }{2}$