Question

The value of $tan40{}^{\circ }+tan20{}^{\circ }+\sqrt{3}tan20{}^{\circ }tan40{}^{\circ }$ is equal to

• A. $\sqrt{12}$
• B. $\frac{1}{\sqrt{3}}$
• C. 1
• D. $\frac{\sqrt{3}}{2}$
• E. $\sqrt{3}$

Answer 1

Answer: (E)

We know that, $tan\text{(}20{}^{\circ }+40{}^{\circ })=\frac{\tan 20{}^{\circ }+\tan 40{}^{\circ }}{1-\tan 20{}^{\circ }\tan 40{}^{\circ }}$
$\Rightarrow$ $\sqrt{3}=\frac{\tan 20{}^{\circ }+\tan 40{}^{\circ }}{1-\tan 20{}^{\circ }\tan 40{}^{\circ }}$
$\Rightarrow$ $\sqrt{3}-\sqrt{3}\tan 20{}^{\circ }\tan 40{}^{\circ }$
$=\tan 20{}^{\circ }+\tan 40{}^{\circ }$
$\Rightarrow$ $\tan 20{}^{\circ }+\tan 40{}^{\circ }+\sqrt{3}\tan 20{}^{\circ }\tan 40{}^{\circ }$
$=\sqrt{3}$