Answer
$\displaystyle \frac{\pi}{2}$
Work Step by Step
$y=\sec^{-1}x$ is the number in $[0, \pi/2) \cup(\pi/2, \pi$] for which $\sec y=x.$
(In terms of cosine, $\quad \displaystyle \cos y=\frac{1}{x}$)
We want an angle (in radians) from $[0, \pi/2$) $\cup(\pi/2, \pi$] such that its cosine approaches 0 (so that its secant grows to $\infty$). This angle is $\displaystyle \frac{\pi}{2}$.
Alternatively (if in doubt), we can reach the same conclusion by observing the graph of $y=\sec x$ (also written as $\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{e}\mathrm{c} x$) when $ x\rightarrow\infty$. See below.
