Q. If $\frac{\tan{\ 3\theta - 1}}{\tan{\ 3\theta + 1}} = \sqrt{3}$, then the general value of $\theta$ is
A) $\frac{n\pi}{3} - \frac{\pi}{12}$
B) $n\pi + \frac{7\pi}{12}$
C) $\frac{n\pi}{3} + \frac{7\pi}{36}$
D) $n\pi + \frac{\pi}{12}$
Explanation
(c)
Given, $\frac{\tan{3\theta - 1}}{\tan{3\theta + 1}} = \sqrt{3}$
$\Rightarrow \ \tan{3\theta - 1 - \sqrt{3}\tan{3\theta - \sqrt{3}} = 0}$
$\Rightarrow \ \tan{3\theta = \frac{1 + \sqrt{3}}{1 - \sqrt{3}} = \tan{(45{^\circ} + 60{^\circ})}}$
$\Rightarrow \ \tan{3\theta = \tan\frac{7\pi}{12}}$
$\Rightarrow \ \ 3\theta = n\pi + \frac{7\pi}{12}$
$\Rightarrow \ \ \ \theta = \frac{n\pi}{3} + \frac{7\pi}{12}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.