## Precalculus (10th Edition)

$x=\sec(y)$, $\geq1$, $0$, $\pi$.
If $\sec{(x)}=y$ then for the inverse function $\sec^{-1}{(y)}=x.$ Hence here it means $x=\sec(y)$, where $|x|\geq1$ and $0\leq y\leq \pi$.