Prove that

Solution:

L.H.S

$\cot 4 x(\sin 5 x+\sin 3 x)$

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

$=\cot 4 x(2 \sin 4 x \cos x)$

$=\frac{\cos 4 x}{\sin 4 x}(2 \sin 4 x \cos x)$

$=2 \cos 4 x \cos x$

R.H.S

$\cot x(\sin 5 x-\sin 3 x)$

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

$=\cot x(2 \cos 4 x \sin x)$

$=\frac{\cos x}{\sin x}(2 \cos 4 x \sin x)$

$=2 \cos 4 x \cos x$

L.H.S=R.H.S

Hence, proved.

Using the formula,

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

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