Q.State the domain and range of $\sec^{-1}x$.

$\sec x$ satisfies $|\sec x|\ge1$ for every $x$ where it is defined (from $\sec^2x=1+\tan^2x\ge1$), so it never takes a value strictly between $-1$ and $1$; hence the domain of $\sec^{-1}x$ is $\mathbb{R}-(-1,1)$, i.e. $x\le-1$ or $x\ge1$. Its range is the principal branch $[0,\pi]$ with the single point $\pi/2$ r …

Use $\sec^2x\ge1$ to justify the excluded domain interval, and remove the point where sec …

Forgetting to remove $\pi/2$ from the range; wrongly making the domain's boundary points $\pm1$ open instead of close …