## University Calculus: Early Transcendentals (3rd Edition)

$sec^{-1}{0}$ is undefined. $\sin^{-1}{\sqrt {2}}$ is undefined.
a) Since $\sec{x}={\frac{1}{cos{x}}}$, $\sec x$ can not be zero and thus $sec^{-1}{0}$ is undefined. b) The value of $\sin^{-1}{\sqrt {2}}$ is not defined as the value of $\sqrt{2}=1.414$, which is greater than 1 and the range of sin is between -1 and 1.