Q. The value of $\tan{40{^\circ} + \tan{20{^\circ} + \sqrt{3}\tan{20{^\circ}\tan{40{^\circ}}}}}$ is equal to
A) $\sqrt{12}$
B) $\frac{1}{\sqrt{3}}$
C) 1
D) $\sqrt{3}$
Explanation
(d)
$\tan{60{^\circ} = \tan{(40{^\circ} + 20{^\circ})}}$
$\Rightarrow \ \sqrt{3} = \frac{\tan{40{^\circ} + \tan{20{^\circ}}}}{1 - \tan{40{^\circ}.\tan{20{^\circ}}}}$
$\Rightarrow \ \sqrt{3} - \sqrt{3}\tan{40{^\circ}\tan{20{^\circ} = \tan{40{^\circ} + \tan{20{^\circ}}}}}$
$\Rightarrow \tan{40{^\circ} + \tan{20{^\circ} + \sqrt{3}\tan{40{^\circ}\tan{20{^\circ} = \sqrt{3}}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.