Prove that $\cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x)$

(A) $L.H.S. = \cot 4x(\sin 5x + \sin 3x)$
$= \frac{\cos 4x}{\sin 4x} \left[ 2 \sin \left( \frac{5x + 3x}{2} \right) \cos \left( \frac{5x - 3x}{2} \right) \right]$
Using the identity $\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$:
$= \left( \frac{\cos 4x}{\sin 4x} \right) [2 \sin 4x \cos x]$
$= 2 \cos 4x \cos x$
$R.H.S. = \cot x(\sin 5x - \sin 3x)$
$= \frac{\cos x}{\sin x} \left[ 2 \cos \left( \frac{5x + 3x}{2} \right) \sin \left( \frac{5x - 3x}{2} \right) \right]$
Using the identity $\sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$:
$= \frac{\cos x}{\sin x} [2 \cos 4x \sin x]$
$= 2 \cos 4x \cos x$
Since $L.H.S. = R.H.S.$,the identity is proved.