What is the General Solution of \(tan \theta\) = \(tan \alpha\) ?

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.

Leave a Comment

Your email address will not be published.