Question

$\frac{d}{dx}\left({\tan }^{-1}\sqrt{\left(\frac{1-\cos x}{1+\cos x}\right)}\right)$

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{-1}{2}$

Answer 1

The correct option is A $\frac{1}{2}$

Let $y={\tan }^{-1}\sqrt{\left(\frac{1-\cos x}{1+\cos x}\right)}$
$\Rightarrow y={\tan }^{-1}\sqrt{\left(\frac{2{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}\right)}$
$\Rightarrow y={\tan }^{-1}\left[\tan \left(\frac{x}{2}\right)\right]=\frac{x}{2}$
$\therefore \frac{dy}{dx}=\frac{1}{2}$