Prove that

Solution:

L.H.S

$=\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}$

$=\frac{1}{2}\left(2 \sin 70^{\circ} \sin 10^{\circ}\right) \sin 50^{\circ} \frac{1}{2}$

$=\frac{1}{4}\left\{\cos \left(70^{\circ}-10^{\circ}\right)-\cos \left(70^{\circ}+10^{\circ}\right)\right\} \sin 50^{\circ}$

$=\frac{1}{4}\left\{\cos 60^{\circ} \sin 50^{\circ}-\cos 80^{\circ} \sin 50^{\circ}\right\}$

$\left.=\frac{1}{4}\left\{\frac{1}{2} \sin 50^{\circ}-\cos 80^{\circ} \sin 50^{\circ}\right\}\right\}$

$=\frac{1}{8}\left\{\sin 50^{\circ}-2 \cos 80^{\circ} \sin 50^{\circ}\right\}$

$=\frac{1}{8}\left\{\sin 50^{\circ}-\left(\sin \left(80^{\circ}+50^{\circ}\right)-\sin \left(80^{\circ}-50^{\circ}\right)\right\}\right.$

$=\frac{1}{8}\left\{\sin 50^{\circ}-\sin 130^{\circ}+\sin 30^{\circ}\right\}$

$=\frac{1}{8}\left\{\sin 50^{\circ}-\sin 130^{\circ}+\frac{1}{2}\right\}$

$=\frac{1}{8}\left\{\sin 50^{\circ}-\sin \left(180^{\circ}-50^{\circ}\right)+\frac{1}{2}\right\}$

$=\frac{1}{8}\left\{\sin 50^{\circ}-\sin 50^{\circ}+\frac{1}{2}\right\}$

$=\frac{1}{16}$

$=$ R.H.S