Prove that

$=\frac{\sin x+\sin 3 x+\sin 5 x}{\cos x+\cos 3 x+\cos 5 x}$

$=\frac{(\sin 5 x+\sin x)+\sin 3 x}{(\cos 5 x+\cos x)+\cos 3 x}$

$=\frac{2 \sin \frac{5 x+x}{2} \cos \frac{5 x-x}{2}+\sin 3 x}{2 \cos \frac{5 x^{2}+x}{2} \cos \frac{5 x-x}{2}+\cos 3 x}$

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

$=\frac{\sin 3 x(2 \cos x+1)}{\cos 3 x(2 \cos x+1)}$

$=\tan 3 x$

Using the formula,

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

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