How do you identify the period and asympotes for y=tantheta?

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Mar 16, 2018

$y = \tan \theta$ has a period of $\pi i$ and vertical asymptotes are at $\theta = \frac{\left(2 n + 1\right) \pi}{2}$

Explanation:

As $\tan \left(\pi + \theta\right) = \tan \theta$, $y = \tan \theta$ has a period of $\pi$.

Further as $\theta \to \frac{\left(2 n + 1\right) \pi}{2}$, from left or right i.e.

${\lim}_{\theta \to {\left(\frac{\left(2 n + 1\right) \pi}{2}\right)}^{+}} \tan \theta = \infty$ and

${\lim}_{\theta \to {\left(\frac{\left(2 n + 1\right) \pi}{2}\right)}^{-}} \tan \theta = - \infty$

Hence vertical asymptotes are at $\theta = \frac{\left(2 n + 1\right) \pi}{2}$

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