Question

$\frac{d}{dx}\left({\tan }^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$ equals If $0<x<\pi$

• A. $1/2$
• B. $1$
• C. $\sec x$
• D. $\csc x$

Answer 1

The correct option is A $1/2$

$\frac{d}{dx}\left({\tan }^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$
$=\frac{\frac{d}{dx}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)}{1+\frac{1-\cos x}{1+\cos x}}=\frac{\frac{\frac{d}{dx}\left(\frac{1-\cos x}{1+\cos x}\right)}{2\sqrt{\frac{1-\cos x}{1+\cos x}}}}{1+\frac{1-\cos x}{1+\cos x}}$
$=(1+\cos x)\frac{d}{dx}(1-\cos x)\frac{-(1-\cos x)\frac{d}{dx}(1+\cos x)}{{(1+\cos x)}^2}\times \frac{1}{2\sqrt{\frac{1-\cos x}{1+\cos x}}(1+\frac{1-\cos x}{1+\cos x})}$

$=\frac{(1+\cos x)\sin x-(1-\cos x)(-\sin x)}{2\sqrt{\frac{1-\cos x}{1+\cos x}}{(1+\cos x)}^2(1+\frac{1-\cos x}{1+\cos x})}$

$=\frac{1}{2}\sqrt{{\tan }^2\frac{x}{2}}\cot \frac{x}{2}=\pm \frac{1}{2}$