Q. If $\tan{(k + 1)\theta = \tan\theta}$, then $\theta$ belongs to the set
A) $\{ n\pi\ :n \in I\}$
B) $\{ n\pi\ /\ 2:n \in I\}$
C) $\{ n\pi\ /\ k\ :n \in I\}$
D) $\{ n\pi\ /\ 2k\ :n \in I\}$
Explanation
(c)
Given, $\tan{(k + 1)\theta = \tan\theta}$
$\Rightarrow \ \ (k + 1)\theta = n\pi + \theta\ \Rightarrow \ \ k\theta = n\pi$
$\Rightarrow \ \ \theta = \frac{n\pi}{k}\ \ \ \ \ \ \ \therefore\ \theta \in \left\{ \frac{n\pi}{k}:n \in I \right\}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.