Question

If $tan\left(k + 1\right)\theta =tan\theta $ , then the set of all the values of $\theta $ is

• A. $\left\{n \pi : n \in I\right\}$
• B. $\left\{\frac{n \pi }{2} : n \in I\right\}$
• C. $\left\{\frac{n \pi }{k} : n \in I\right\}$
• D. $\left\{\frac{n \pi }{4} : n \in I\right\}$

Answer 1

Answer: (C)

As we know the general solution of the equation
$tanx=tan\alpha $ is given by $x=n\pi +\alpha ,n\in I$
Hence, the solution of the equation, $tan\left(k + 1\right)\theta =tan\theta $
is given by $\left(k + 1\right)\theta =n\pi +\theta ,n\in I.$
$\Rightarrow k\theta =n\pi ,n\in I$ .
$\therefore \theta \in \frac{n \pi }{k},n\in I$