# How do you find a vertical asymptote for y = sec(x)?

The vertical asymptotes of ##y=secx## are

##x={(2n+1)pi}/2##, where ##n## is any integer,

which look like this (in red).

Let us look at some details.

##y=secx=1/{cosx}##

In order to have a vertical asymptote, the (one-sided) limit has to go to either ##infty## or ##-infty##, which happens when the denominator becomes zero there.

So, by solving

##cosx=0##

##Rightarrow x=pm pi/2, pm{3pi}/2, pm{5pi}/2, …##

##Rightarrow x=pi/2+npi={(2n+1)pi}/2##, where ##n## is any integer.

Hence, the vertical asymptotes are

##x={(2n+1)pi}/2##, where ##n## is any integer.