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

(N/A) We use the trigonometric identities:
$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
Applying these to the $L.H.S.$:
$L.H.S. = \frac{2 \sin \left( \frac{5x+3x}{2} \right) \cos \left( \frac{5x-3x}{2} \right)}{2 \cos \left( \frac{5x+3x}{2} \right) \cos \left( \frac{5x-3x}{2} \right)}$
Simplifying the terms:
$= \frac{2 \sin(4x) \cos(x)}{2 \cos(4x) \cos(x)}$
Canceling the common terms $2$ and $\cos(x)$:
$= \frac{\sin(4x)}{\cos(4x)} = \tan(4x) = R.H.S.$
Hence,the identity is proved.