Answer
$\dfrac{2\sqrt 3}{3}$
Work Step by Step
When $P$ is the period of $f(x)$ then we have
$ f(x+p)=f(x)$
Here, we have $ \theta= \dfrac{11 \pi}{6}$ lies in the fourth quadrant thus, the angle is in the interval $[0, 2 \pi]$.
But $\theta'=2 \pi-\dfrac{11 \pi}{6}=\dfrac{\pi}{6}$
and $\sec \theta'=\sec \dfrac{\pi}{6}=\dfrac{1}{\cos \dfrac{\pi}{6}}$
$\dfrac{1}{\cos \dfrac{\pi}{6}}=\dfrac{1}{\sqrt 3/2}$
or, $=\dfrac{2}{\sqrt 3}$
or, $=\dfrac{2\sqrt 3}{3}$
