Answer
$\dfrac {2}{\sqrt {3}}$
Work Step by Step
Period of $sec$ function is $2\pi$
It means
$sec\left( \alpha \right) =sec\left( \alpha +2\pi k\right) =sec\left( \alpha -2\pi k\right) $
$k$ is integer number
$sec\left( \dfrac {13\pi }{6}\right) =sec\left( 2\pi +\dfrac {\pi }{6}\right) =sec\dfrac {\pi }{6}=\dfrac {1}{\cos \dfrac {\pi }{6}}=\dfrac {2}{\sqrt {3}}$
