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The value of $\tan 20 ° + 4 \sin 20 °$ is

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$\frac{2}{\sqrt{3}}$

$\sqrt{3}$

$2 \sqrt{3}$

$\frac{\sqrt{3}}{2}$

answer is A.

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Detailed Solution

$\tan 20 ° + 4 \sin 20 °$

$= \frac{\sin 20 ° + 4 \sin 20 ° \cos 20 °}{\cos 20 °}$

$= \frac{1}{\cos 20 °} [\sin 20 ° + 2 \sin 40 °]$

$= \frac{1}{\cos 20 °} [\sin 20 ° + \sin 40 ° + \sin 40 °]$

$= \frac{1}{\cos 20 °} [2 \sin (30 °) . \cos (10 °) + \sin 40 °]$

$= \frac{1}{\cos 20 °} [\cos 10 ° + \sin 40 °]$

$= \frac{1}{\cos 20 °} [\cos 10 ° + \cos 50 °]$

$= \frac{1}{\cos 20 °} [2 \cos (30 °) . \cos (20 °)]$

$= 2 \times \cos 30 °$

$= 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$

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