Question

$\sin 10^{\circ }\sin 50^{\circ }\sin 60^{\circ }\sin 70^{\circ }=m$ and $\tan 20^{\circ }\tan 40^{\circ }\tan 60^{\circ }\tan 80^{\circ }=n$, then $\frac{n}{m}=$

• A. $\frac{3\sqrt{3}}{16}$
• B. $16\sqrt{3}$
• C. $\frac{16}{\sqrt{3}}$
• D. $8\sqrt{3}$

Answer 1

Answer: (B)

$\sin 10^{\circ }\sin 50^{\circ }\sin 60^{\circ }\sin 70^{\circ }=m$ and $\tan 20^{\circ }\tan 40^{\circ }\tan 60^{\circ }\tan 80^{\circ }=n$
Then, $\frac{n}{m}$ =?
$m=\frac{\sqrt{3}}{2}\cdot \frac{1}{2}\left[2\sin 10^{\circ }\sin 50^{\circ }\right]\cdot \sin 70^{\circ }$
$\Rightarrow m=\frac{\sqrt{3}}{4}\left[\cos 40^{\circ }-\cos 60^{\circ }\right]\sin 70^{\circ }$
$=\frac{\sqrt{3}}{4}\left[\cos 40^{\circ }-\frac{1}{2}\right]\cdot \sin 70^{\circ }$
$=\frac{\sqrt{3}}{4}\cos 40^{\circ }\sin 70^{\circ }-\frac{\sqrt{3}}{4}\frac{1}{2}\sin 70^{\circ }$
$=\frac{1}{2}\frac{\sqrt{3}}{4}\left(2\sin 50^{\circ }\sin 70^{\circ }\right)-\frac{1}{2}\sin 70^{\circ }\cdot \frac{\sqrt{3}}{4}$
$=\frac{\sqrt{3}}{8}\left[\cos 20^{\circ }-\cos 120^{\circ }\right]-\frac{\sqrt{3}}{8}\sin 70^{\circ }$
$=\frac{\sqrt{3}}{8}\cos 20^{\circ }+\frac{\sqrt{3}}{8}\frac{1}{2}-\frac{\sqrt{3}}{8}\sin 70^{\circ }=\frac{\sqrt{3}}{16}=m$
$=\tan 60^{\circ }\cdot \tan 20^{\circ }\cdot \tan 40^{\circ }\cdot \tan 80^{\circ }$
$=\tan 60^{\circ }\cdot \tan 20^{\circ }\cdot \tan \left(60^{\circ }-20^{\circ }\right)\cdot \tan \left(60^{\circ }+20^{\circ }\right)$
$\Rightarrow \tan 60^{\circ }\cdot \tan 3(20)$
$\Rightarrow \tan 60^{\circ }\cdot \tan 60=n$
$n=3$
$\frac{n}{m}=\frac{3\times 16}{\sqrt{3}}$
$\Rightarrow 16\sqrt{3}$