Trigonometry! #147

We know that $\pi = 180^{\circ}$ From the given equation, after substituting $\pi$ with $180^{\circ}$ , we get

$= \sin 10^{\circ}\sin50^{\circ}\sin70^{\circ}$

$\sin10^{\circ}[\sin(60^{\circ}-10^{\circ}) \sin(60^{\circ}+10^{\circ})]$

$\sin10^{\circ}[ \sin^2 60^{\circ} - sin^2 10^{\circ}]$

$\sin10^{\circ}[ \frac{3}{4} - \sin^2 10^{\circ}]$

Taking $\frac{1}{4}$ common, we get

$\frac{1}{4}[ 3\sin10^{\circ} - 4\sin^3 10^{\circ}]$

$\frac{1}{4}[ \sin(3\times10^{\circ})]$

$\frac{1}{4}[\sin30^{\circ}]$

$\frac{1}{4}\times\frac{1}{2}$

$\frac{1}{8}$

Hence, reciprocal= $8$