Question

Prove that: $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$

Answer 1

$\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$
$L.H.S=\frac{\sin x-\sin y}{\cos x+\cos y}$
$\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$
$\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$
$=\frac{2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)}{2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}=\frac{\sin \left(\frac{x-y}{2}\right)}{\cos \left(\frac{x-y}{2}\right)}$
$=\tan \frac{x-y}{2}=R.H.S.$
Hence proved.