# How do you find vertical asymptote of tangent?

Aug 27, 2014

I assume that you are asking about the tangent function, so $\tan \theta$. The vertical asymptotes occur at the NPV's: $\theta = \frac{\pi}{2} + n \pi , n \in \mathbb{Z}$.

Recall that $\tan$ has an identity: $\tan \theta = \frac{y}{x} = \frac{\sin \theta}{\cos \theta}$. This means that we will have NPV's when $\cos \theta = 0$, that is, the denominator equals 0.

$\cos \theta = 0$ when $\theta = \frac{\pi}{2}$ and $\theta = \frac{3 \pi}{2}$ for the Principal Angles. Normally, we have 2 solutions, but the spacing between these 2 angles are the same, so we have a single solution,

$\theta = \frac{\pi}{2} + n \pi , n \in \mathbb{Z}$ in radians or
$\theta = 90 + 180 n , n \in \mathbb{Z}$ for degrees.

To find the vertical asymptote of ANY function, we look for when the denominator is 0.