Question

If ${\tan }^{-1}\left(\sqrt{\cos 2x}\right)-{\cot }^{-1}\left(\sqrt{\cos 2x}\right)=x,$ then which of the following options is always CORRECT?

• A. $\sin x={\tan }^{2}x$
• B. $\sin x=-{\tan }^{2}x$
• C. $\sin x={\cot }^{2}x$
• D. $\sin x=-{\cot }^{2}x$

Answer 1

The correct option is B $\sin x=-{\tan }^{2}x$

${\tan }^{-1}\left(\sqrt{\cos 2x}\right)-{\cot }^{-1}\left(\sqrt{\cos 2x}\right)=x$
$\Rightarrow 2{\tan }^{-1}\left(\sqrt{\cos 2x}\right)=x+\frac{\pi }{2}$
$(\because {\tan }^{-1}A+{\cot }^{-1}A=\frac{\pi }{2})$
$\Rightarrow \sqrt{\cos 2x}=\tan (\frac{x}{2}+\frac{\pi }{4})$
$\Rightarrow \sqrt{\cos 2x}=\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}$

Squaring both sides, we get
$\cos 2x=\frac{1+2\cos \frac{x}{2}\sin \frac{x}{2}}{1-2\cos \frac{x}{2}\sin \frac{x}{2}}$
$\Rightarrow \frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{1+\sin x}{1-\sin x}$

Applying componendo and dividendo,
$\frac{2}{-2{\tan }^{2}x}=\frac{2}{2\sin x}$
$\Rightarrow \sin x=-{\tan }^{2}x$