Prove that $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \left(\frac{x-y}{2}\right)$

We use the following trigonometric identities:
$\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$
$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
Applying these to the $L.H.S.$:
$L.H.S. = \frac{\sin x-\sin y}{\cos x+\cos y}$
$= \frac{2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}$
Canceling the common terms $2 \cos \left(\frac{x+y}{2}\right)$:
$= \frac{\sin \left(\frac{x-y}{2}\right)}{\cos \left(\frac{x-y}{2}\right)}$
$= \tan \left(\frac{x-y}{2}\right) = R.H.S.$