Question

Solve $\tan {20}^{\circ }+\tan {40}^{\circ }\sqrt{3}\tan {20}^{\circ }\tan {40}^{\circ }$

Answer 1

The question must be

Solve:$\tan {20}^{\circ }+\tan {40}^{\circ }+\sqrt{3}\tan {20}^{\circ }\tan {40}^{\circ }$

Consider $\tan {20}^{\circ }+\tan {40}^{\circ }$

We know that $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$=\tan ({20}^{\circ }+{40}^{\circ })(1-\tan {20}^{\circ }\tan {40}^{\circ })+\sqrt{3}\tan {20}^{\circ }\tan {40}^{\circ }$

$=\tan {60}^{\circ }(1-\tan {20}^{\circ }\tan {40}^{\circ })+\sqrt{3}\tan {20}^{\circ }\tan {40}^{\circ }$

$=\sqrt{3}-\sqrt{3}\tan {20}^{\circ }\tan {40}^{\circ }+\sqrt{3}\tan {20}^{\circ }\tan {40}^{\circ }$

$=\sqrt{3}$