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