Question

Prove that
${\tan }^{-1}\left(\frac{\sin x}{1+\cos x}\right),-\pi <x<\pi$

Answer 1

Prove that $ta{n}^{-1}\left(\frac{sinx}{1+cosx}\right)$

$ta{n}^{-1}\left(\frac{sinx}{1+cosx}\right)=ta{n}^{-1}\left(\frac{2sinx/2.cosx/2}{2co{s}^{2}x/2}\right)$

$=ta{n}^{-1}\left(tanx/2\right)$

$=\frac{x}{2}$

$(\because -\pi <x<\pi )$

$(-\frac{\pi }{2}<\frac{x}{2}<\frac{\pi }{2})$