Prove that $\frac{\sin x+\sin 3x}{\cos x+\cos 3x} = \tan 2x$.

(N/A) We use the sum-to-product identities:
$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
Applying these to the $L.H.S.$:
$L.H.S. = \frac{\sin x + \sin 3x}{\cos x + \cos 3x}$
$= \frac{2 \sin \left( \frac{x+3x}{2} \right) \cos \left( \frac{x-3x}{2} \right)}{2 \cos \left( \frac{x+3x}{2} \right) \cos \left( \frac{x-3x}{2} \right)}$
$= \frac{\sin(2x) \cos(-x)}{\cos(2x) \cos(-x)}$
Since $\cos(-x) = \cos(x)$,we can cancel the $\cos(-x)$ terms:
$= \frac{\sin 2x}{\cos 2x} = \tan 2x = R.H.S.$