Answer
$2$
Work Step by Step
The range of $\cos^{-1}{x}$ is $[0, \pi]$.
Here, $\cos^{-1}{(\dfrac{1}{2})}=\dfrac{ \pi}{3}$, because $\cos{(\dfrac{\pi}{3})}=-\dfrac{1}{2}$ and $\dfrac{\pi}{3}$ lies in the range of $\cos^{-1}{x}$ .
Therefore, we have:
$\sec [\cos^{-1}{(\dfrac{1}{2})}]=\sec ({\dfrac{ \pi}{3}})=2$