Answer
$\dfrac{2 \sqrt 3}{3}$
Work Step by Step
The reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps:
a) Quadrant- I: $\theta $
b) Quadrant- II: $180^{\circ}-\theta $
c) Quadrant -III: $\theta - 180^o $
d) Quadrant -IV: $360^{\circ}-\theta $
The reference angle is equal to $ 2\pi- \dfrac{11\pi}{6}=\dfrac{\pi}{6}$
Since, $\sec (\dfrac{\pi}{6})=\dfrac{2 \sqrt 3}{3}$
So, $\sec \dfrac{11\pi}{6}=\dfrac{2 \sqrt 3}{3}$ because $\theta $ lies in Quadrant-IV.
