Prove that

$\frac{\sin x+\sin 3 x}{\cos x-\cos 3 x}$

$=\frac{2 \sin \frac{3 x+x}{2} \cos \frac{3 x-x}{2}}{-2 \sin \frac{x+3 x}{2} \sin \frac{x-3 x}{2}}$

$=\frac{2 \sin \frac{4 x}{2} \cos \frac{2 x}{2}}{2 \sin \frac{4 x}{2} \sin \frac{2 x}{2}}$

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

$=\cot x$

Using the formula,

$\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$

$\cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$