Question:
$\tan 20^{\circ}+\tan 40^{\circ}+\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}$ is equal to
(a) $\frac{\sqrt{3}}{4}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\sqrt{3}$
(d) 1
Solution:
(c) $\sqrt{3}$
$\tan 20^{\circ}+\tan 40^{\circ}+\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}$
$=\tan 60^{\circ}\left(1-\tan 20^{\circ} \tan 40^{\circ}\right)+\tan 60^{\circ} \tan 20^{\circ} \tan 40^{\circ}$
$\left[\right.$ Using $\tan 60^{\circ}=\frac{\tan 20+\tan 40}{1-\tan 20 \tan 40}$ and $\left.\tan 60^{\circ}=\sqrt{3}\right]$
$=\tan 60^{\circ}-\tan 60^{\circ} \tan 20^{\circ} \tan 40^{\circ}+\tan 60^{\circ} \tan 20^{\circ} \tan 40^{\circ}$
$=\tan 60^{\circ}$
$=\sqrt{3}$