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  4. sin 10° sin 50° sin 60° sin 70°=m and tan 20° tan 40° tan 60° tan 80°=n, then (n/m)=

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

TS EAMCET 2018

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, $\frac{n}{m}$ =?
$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}$