Prove that $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$

It is known that

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

$\therefore$ $L.H.S.$ $=\sin ^{2} 6 x-\sin ^{2} 4 x$

$=(\sin 6 x+\sin 4 x)(\sin 6 x-\sin 4 x)$

$=\left[2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right)\right]\left[2 \cos \left(\frac{6 x+4 x}{2}\right) \cdot \sin \left(\frac{6 x-4 x}{2}\right)\right]$

$=(2 \sin 5 x \cos x)(2 \cos 5 x \sin x)=(2 \sin 5 x \cos 5 x)(2 \sin x \cos x)$

$=\sin 10 x \sin 2 x$

$= R . H.S$