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