Answer
$D=(−∞,-1]\cup[1,∞)$
$R=\left[0,\dfrac{\pi}{2}\right)\cup\left(\dfrac{\pi}{2},\pi\right]$
Work Step by Step
Consider the function:
$f(x)=\sec x$
For the inverse to exist, we must restrict the domain of $f$ so that the function is one-to-one:
$D_f=\left[0,\dfrac{\pi}{2}\right)\cup\left(\dfrac{\pi}{2},\pi\right]$
$R_f=(−∞,-1]\cup[1,∞)$
The domain and range of the inverse $f^{−1}(x)=\sec^{−1}(x)$ are:
$D_{f^{−1}}=(−∞,-1]\cup[1,∞)$
$R_{f^{−1}}=\left[0,\dfrac{\pi}{2}\right)\cup\left(\dfrac{\pi}{2},\pi\right]$
