Answer
$\frac{2\sqrt 3}{3}$
Work Step by Step
$\sec(\tan^{-1}\frac{\sqrt 3}{3})$ is the question.
Tangent value is $\frac{\sqrt 3}{3}$ when it is at $\frac{\pi}{6}$
This is the only possible answer because tangent has to be in quadrants 1 or 4.
At $\frac{\pi}{6}$, the cosine value is $\frac{\sqrt 3}{2}$.
Secant is $1/\cos$ so replacing cos with $\frac{\sqrt 3}{2}$, and simplifying it, we get $\frac{2\sqrt 3}{3}$.
