Q. If $\tan{\theta + \tan{2\theta + \sqrt{3}\tan{\theta\tan{2\theta = \sqrt{3}}}}}$, then
A) $\theta = \frac{(6n + 1)\pi}{18},\forall\ n \in I$
B) $\theta = \frac{(6n + 1)\pi}{9},\forall\ n \in I$
C) $\theta = \frac{(3n + 1)\pi}{9},\forall\ n \in I$
D) None of these
Explanation
(c)
$\because\ \tan{\theta + \tan{2\theta + \sqrt{3}\tan{\theta\tan{2\theta = \sqrt{3}}}}}$
$\Rightarrow \ \tan{\theta + \tan{2\theta = \sqrt{3}(1 - \tan{\theta\tan{2\theta)}}}}$
$\Rightarrow \ \frac{\tan{\theta + \tan 2}\theta}{1 - \tan{\theta\tan{2\theta}}} = \sqrt{3}\ \ \Rightarrow \ \tan{3\theta = \tan\frac{\pi}{3}}$
$\Rightarrow \ 3\theta = n\pi + \frac{\pi}{3}\ \ \Rightarrow \ \theta = \frac{(3n + 1)\pi}{9},\ n \in I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.