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.
