Question

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

• A. $1/2$
• B. $-1$
• C. $1/4$
• D. $-1/2$

Answer 1

The correct option is A $1/2$

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