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TS EAMCET 4 May 2018 Shift 2 Solved Paper

Section: Mathematics

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Question : 20 of 160

Marks: +1, -0

\sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ}=m

\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}=n

\frac{n}{m}=

$3\sqrt{3}.16$
$16\sqrt{3}$
$\frac{16}{\sqrt{3}}$
$8 \sqrt{3}$

Solution:

The given equations are,

\sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ}=m

\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}=n

Then,

\frac{\sqrt{3}}{2} \left( \frac{1}{2} \right) \left[ 2 \sin 10^{\circ} \sin 50 \right] \sin 70^{\circ}=m

\frac{\sqrt{3}}{4} \left[ \cos 40^{\circ} - \cos 60^{\circ} \right] \sin 70^{\circ}=m

\frac{\sqrt{3}}{8} \left[ \cos 20^{\circ} - \cos 120^{\circ} \right]

-\frac{\sqrt{3}}{8} \sin 70^{\circ}=m

m=\frac{\sqrt{3}}{16}

Now,

\tan 60^{\circ} \tan 20^{\circ} \tan 40^{\circ} \tan 80^{\circ}=n

\tan 60^{\circ} \tan 20^{\circ} \tan (60^{\circ}-20^{\circ}) \tan (60^{\circ}+20^{\circ})=n

\tan 60^{\circ} \tan 3 (20^{\circ})=n

n=3

This implies,

\frac{n}{m} = \frac{3 \times 16}{\sqrt{3}}

=16 \sqrt{3}

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