Q60 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics
$\sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ}=m$ and $\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}=n$, then $\frac{n}{m}=$
- A$\frac{3 \sqrt{3}}{16}$
- Correct answerB$16 \sqrt{3}$
- C$\frac{16}{\sqrt{3}}$
- D$8 \sqrt{3}$
Solution
$\sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ}=m$ and $\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}=n$ Then, $\frac{n}{m}=?$
$$
\begin{aligned}
m & =\frac{\sqrt{3}}{2} \cdot \frac{1}{2}\left[2 \sin 10^{\circ} \sin 50^{\circ}\right] \cdot \sin 70^{\circ} \\
\Rightarrow m & =\frac{\sqrt{3}}{4}\left[\cos 40^{\circ}-\cos 60^{\circ}\right] \sin 70^{\circ} \\
& =\frac{\sqrt{3}}{4}\left[\cos 40^{\circ}-\frac{1}{2}\right] \cdot \sin 70^{\circ} \\
& =\frac{\sqrt{3}}{4} \cos 40^{\circ} \sin 70^{\circ}-\frac{\sqrt{3}}{4} \frac{1}{2} \sin 70^{\circ}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{1}{2} \frac{\sqrt{3}}{4}\left(2 \sin 50^{\circ} \sin 70^{\circ}\right)-\frac{1}{2} \sin 70^{\circ} \cdot \frac{\sqrt{3}}{4} \\
& =\frac{\sqrt{3}}{8}\left[\cos 20^{\circ}-\cos 120^{\circ}\right]-\frac{\sqrt{3}}{8} \sin 70^{\circ} \\
& =\frac{\sqrt{3}}{8} \cos 20^{\circ}+\frac{\sqrt{3}}{8} \frac{1}{2}-\frac{\sqrt{3}}{8} \sin 70^{\circ}=\frac{\sqrt{3}}{16}=m \\
& =\tan 60^{\circ} \cdot \tan 20^{\circ} \cdot \tan 40^{\circ} \cdot \tan 80^{\circ} \\
& =\tan 60^{\circ} \cdot \tan 20^{\circ} \cdot \tan \left(60^{\circ}-20^{\circ}\right) \cdot \tan \left(60^{\circ}+20^{\circ}\right) \\
& \Rightarrow \tan 60^{\circ} \cdot \tan 3(20) \Rightarrow \tan 60^{\circ} \cdot \tan 60=n \\
& \qquad n=3 \\
& \quad \frac{n}{m}=\frac{3 \times 16}{\sqrt{3}} \Rightarrow 16 \sqrt{3}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.