Prove that $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$

solution:

It is known that

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

$\therefore$ L.H.S. $=\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}$

$=\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cdot \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cdot \cos \left(\frac{5 x-3 x}{2}\right)}$

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

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

$=\tan 4 \mathrm{x}=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S}$