Question

Differentiate

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

Answer 1

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

This function is defined for all real numbers where cos x ≠ 1

$$\begin{gathered} f\left(x\right)={\tan }^{-1}\left(\frac{\sin x}{1+\cos x}\right) \\ \Rightarrow f\left(x\right)={\tan }^{-1}\left[\frac{2\sin \left({\displaystyle \frac{x}{2}}\right)\cos \left({\displaystyle \frac{x}{2}}\right)}{2{\cos }^2\left({\displaystyle \frac{x}{2}}\right)}\right] \\ \Rightarrow f\left(x\right)={\tan }^{-1}\left[\tan \left(\frac{x}{2}\right)\right]=\frac{x}{2} \\ \mathrm{Thus},f\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{x}{2}\right)=\frac{1}{2} \end{gathered}$$