Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

L.H.S $=\cot 4 x(\sin 5 x+\sin 3 x)$

$=\frac{\cos 4 x}{\sin 4 x}\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right]$

$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$

$=\left(\frac{\cos 4 x}{\sin 4 x}\right)[2 \sin 4 x \cos x]$

$=2 \cos 4 x \cos x$

R.H.S. $=\cot x(\sin 5 x-\sin 3 x)$

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

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

$=\frac{\cos x}{\sin x}[2 \cos 4 x \sin x]$

$=2 \cos 4 x \cdot \cos x$

L.H.S. = R.H.S.