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

solution:

It is known that

$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right), \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$

$\therefore$ L.H.S. $=\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$

$=\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}$

$=\frac{\sin 2 x}{\cos 2 x}$

$=\tan 2 \mathrm{x}$

$=$ R.H.S