Q. The solution of $\tan{2\theta\tan{\theta = 1}}$ is22
A) $\frac{\pi}{3}$
B) $(6n \pm 1)\frac{\pi}{6}$
C) $(4n \pm 1)\frac{\pi}{6}$
D) $(2n + \pi)\frac{\pi}{6}$
Explanation
(b)
Given, $\tan{2\theta\tan{\theta = 1}}$
$\therefore\ \ \ \frac{2\tan^{2}\theta}{1 - \tan^{2}\theta} = 1\ \ \ \Rightarrow \ \tan^{2}{\theta = \frac{1}{3}}$
$\Rightarrow \ \tan^{2}{\theta = \tan^{2}{\frac{\pi}{6}\ \ \ \Rightarrow \ \ \ \ \theta = n\pi \pm \frac{\pi}{6}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.