Prove that:

$\mathrm{LHS}=\frac{1}{4}\left(2 \sin 6^{\circ} \sin 66^{\circ}\right)\left(2 \sin 42^{\circ} \sin 78^{\circ}\right)$

$=\frac{1}{4}\left(\cos 60^{\circ}-\cos 72^{\circ}\right)\left(\cos 36^{\circ}-\cos 120^{\circ}\right)$

$[2 \sin A \sin B=\cos (A-B)-\cos (A+B)]$

$=\frac{1}{4}\left(\frac{1}{2}-\sin 18^{\circ}\right)\left(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right)$

$=\frac{1}{4}\left(\frac{1}{2}-\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right)$

$=\frac{1}{4}\left(\frac{2-\sqrt{5}+1}{4}\right)\left(\frac{\sqrt{5}+1+2}{4}\right)$

$=\frac{1}{64}(3-\sqrt{5})(3+\sqrt{5})$

$=\frac{1}{64}(9-5)=\frac{1}{16}$

= RHS

Hence proved.