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. $=\sin ^{2} 6 x-\sin ^{2} 4 x$

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

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

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

$=\sin 10 x \sin 2 x$

$=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S} .$