The value of $\tan (20^{\circ })+\tan (40^{\circ })+\sqrt{3}\tan (20^{\circ })\tan (40^{\circ })$ is equal to

Solution

1. Recall the tangent addition formula: $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$.
2. Set $A=20^{\circ }$ and $B=40^{\circ }$ to get $\tan (60^{\circ })=\frac{\tan 20^{\circ }+\tan 40^{\circ }}{1-\tan 20^{\circ }\tan 40^{\circ }}$.
3. Since $\tan (60^{\circ })=\sqrt{3}$, we have $\sqrt{3}(1-\tan 20^{\circ }\tan 40^{\circ })=\tan 20^{\circ }+\tan 40^{\circ }$.
4. Distribute $\sqrt{3}$ to obtain $\sqrt{3}-\sqrt{3}\tan 20^{\circ }\tan 40^{\circ }=\tan 20^{\circ }+\tan 40^{\circ }$.
5. Rearrange the terms to find $\tan 20^{\circ }+\tan 40^{\circ }+\sqrt{3}\tan 20^{\circ }\tan 40^{\circ }=\sqrt{3}$.
6. Hence the answer is (A).