Mathematics - Trigonometric Equation Question with Solution | TestHub

$\begin{array}{rl}\sqrt{3}& =\tan 60^{\circ }=\tan \left(40^{\circ }+20^{\circ }\right)\\ & =\frac{\tan 40^{\circ }+\tan 20^{\circ }}{1-\tan 40^{\circ }\tan 20^{\circ }}\\ \therefore & \sqrt{3}-\sqrt{3}\tan 40^{\circ }\tan 20^{\circ }\\ & =\tan 40^{\circ }+\tan 20^{\circ }\end{array}$ Hence $\tan 40^{\circ }+\tan 20^{\circ }+\sqrt{3}\tan 40^{\circ }\tan 20^{\circ }$ $=\sqrt{3}$