Prove that

$\frac{\sin 7 x-\sin 5 x}{\cos 7 x+\cos 5 x}$

$=\frac{2 \cos \frac{7 x+5 x}{2} \sin \frac{7 x_{-} 5 x}{2}}{2 \cos \frac{7 x+5 x}{2} \cos \frac{7 x-5 x}{2}}$

$=\frac{2 \cos 6 x \sin x}{2 \cos 6 x \cos x}$

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

$=\tan x$

Using the formula,

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

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