Q. If $\tan{\theta + \tan{\left( \frac{\pi}{3} + \theta \right) + \tan{\left( - \frac{\pi}{3} + \theta \right) = a\tan{3\theta,}}}}$ then $a$ is equal to

A) $\frac{1}{3}$
B) 1
C) 3
D) None of these
Explanation

(c)
$\tan{\theta + \tan\left( \frac{\pi}{3} + \theta \right) + \tan{\left( - \frac{\pi}{3} + \theta \right) = a\tan{3\theta}}}$
$\Rightarrow \tan{\theta + \frac{\sqrt{3} + \tan\theta}{1 - \sqrt{3}\tan\theta} + \frac{\tan{\theta - \sqrt{3}}}{1 + \sqrt{3}\tan\theta} = {a\ tan}{3\theta}}$
$\Rightarrow \tan{\theta + \frac{8\tan\theta}{1 - 3\tan^{2}\theta} = a\tan{3\theta}}$
$\Rightarrow \frac{3(3\tan{\theta - \tan^{3}{\theta)}}}{1 - 3\tan^{2}\theta} = a\tan{3\theta}$
$\Rightarrow 3\tan{3\theta = a\tan{3\theta}}$
$\Rightarrow a = 3$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.