Solution :
The general solution of \(tan \theta\) = \(tan \alpha\) is given by \(\theta\) = \(n\pi + \alpha\), n \(\in\) Z.
Proof :
We have, \(tan \theta\) = \(tan \alpha\)
\(\implies\) \(sin \theta\over cos \theta\) = \(sin \alpha\over cos \alpha\)
\(\implies\) \(sin \theta cos \alpha\) – \(cos \theta sin \alpha\) = 0
\(\implies\) \(sin (\theta – \alpha)\) = 0
\(\implies\) \(\theta – \alpha\) = \(n\pi\), n \(\in\) Z
\(\implies\) \(\theta\) = \(n\pi + \alpha\), n \(\in\) Z
Remark : Since \(tan \theta\) = \(tan \alpha\) is equivalent to \(cot \theta\) = \(cot \alpha\). So, general solutions of \(cot \theta\) = \(cot \alpha\) and \(tan \theta\) = \(tan \alpha\) are same.
