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

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A

$\left\{\right.n\pi:n\in I\left.\right\}$

B

$\left\{\frac{\text{n} \pi }{2} \text{:} \text{n} ∈ \textit{I}\right\}$

C

$\left\{\frac{\text{n} \pi }{\text{k}} \text{:} \text{n} \in \textit{I}\right\}$

D

$\left\{\frac{\text{n} \pi }{4} \text{:} \text{n} ∈ \textit{I}\right\}$

Solution:

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

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