Question

$\tan {20}^{\circ }+\tan {40}^{\circ }+\sqrt{3}\tan {20}^{\circ }\tan {40}^{\circ }$ is equal to

• A. $\frac{1}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. $-\frac{1}{\sqrt{3}}$
• D. $-\sqrt{3}$

Answer 1

The correct option is B $\sqrt{3}$

${40}^{\circ }+{20}^{\circ }={60}^{\circ }$

$\Rightarrow \tan ({40}^{\circ }+{20}^{\circ })=\tan {60}^{\circ }$

we know that $\tan (\alpha +\beta )=\frac{\tan \left(\alpha \right)+\tan \left(\beta \right)}{1-\tan \left(\alpha \right)\tan \left(\beta \right)}$

$\Rightarrow \frac{\tan {40}^{\circ }+\tan {20}^{\circ }}{1-\tan {40}^{\circ }\tan {20}^{\circ }}=\sqrt{3}$

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

$\Rightarrow \tan {40}^{\circ }+\tan {20}^{\circ }=\sqrt{3}-\sqrt{3}\tan {40}^{\circ }\tan {20}^{\circ }$
$\therefore \tan {40}^{\circ }+\tan {20}^{\circ }+\sqrt{3}\tan {40}^{\circ }\tan {20}^{\circ }=\sqrt{3}$
Ans: B