#### Answer

$D$

#### Work Step by Step

The secant function is undefined when $x=-\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, ...$
Thus, its graph has the vertical asymptotes $x=-\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2},...$.
RECALL:
The graph of the function $y=\sec{(x-d)}$ involves a phase (horizontal) shift of the parent function $y=\sec{x}$.
The shift is to the right when $d\gt0$ and to the left when $d\lt0$.
The given function has $d=\frac{\pi}{2}$ so it involves a phase shift of $\frac{\pi}{2}$ to the right.
Thus, its vertical asymptotes are $x=0, \pi, 2\pi, ...$
The only possible graph among the choices are the ones in Options B and D.
Note that when $x=\frac{\pi}{2}$ , the value of $y=\sec{(x-\frac{\pi}{2})}$ is $1$.
Only the graph in Option $C$ satisfies this.
Thus, the graph of $y=\sec{(x-\frac{\pi}{2})}$ must be the one in Option $D$.