## Precalculus (6th Edition)

$\sec^{-1}0$ does not exist
$y=\sec^{-1}x =$arcsec$x$ Domain: $(-\infty, -1]\cup[1, \infty)$ Range: $[0,\displaystyle \frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$ Quadrants (unit circle): I and II -------- 0 is not in the domain of $y=\sec^{-1}x,$ because 0 is not in the range of $\sec x$. (There is no y from $[0,\displaystyle \frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$ such that $\displaystyle \sec y=\frac{1}{\cos y}=0)$ $\sec^{-1}0$ does not exist