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 $\sec y=x.$
(If you prefer to think in terms of cosine, then $\displaystyle \cos y=\frac{1}{x}$)
When cosine approaches zero from the left, the secant function "runs off" toward $-\infty.$
The angle (in radians) from $[0, \pi/2) \cup(\pi/2, \pi]$ for which this happens is $\displaystyle \frac{\pi}{2}$ (when approached from the right).
Alternatively, 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.
