Answer
$\sec^{-1}0$ does not exist
Work Step by Step
$y=\sec^{-1}x =$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
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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