Prove that:

To Prove: $\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)=\frac{x}{2}$

Formula Used:

1) $\sin A=2 \times \sin \frac{A}{2} \times \cos \frac{A}{2}$

2) $1+\cos A=2 \cos ^{2} \frac{A}{2}$

Proof:

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

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

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

$=\tan ^{-1}\left(\tan \frac{x}{2}\right)$

$=\frac{\mathrm{x}}{2}$

$=\mathrm{RHS}$

Therefore $\mathrm{LHS}=\mathrm{RHS}$

Hence proved.