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