Answer
$-2$
Work Step by Step
Period of $sec$ function is $2\pi$
It means
$sec\left( \alpha \right) =sec\left( \alpha +2\pi n\right) =sec\left( \alpha -2\pi n\right) $
$n$ is integer number
$sec\left( \dfrac {22\pi }{3}\right) =sec\left( 6\pi +\dfrac {4\pi }{3}\right) =sec\dfrac {4\pi }{3}=\dfrac {1}{\cos \dfrac {4\pi }{3}}=\dfrac {1}{\cos \left( \pi +\dfrac {\pi }{3}\right) }=\dfrac {1}{\cos \pi \cos \dfrac {\pi }{3}-\sin \pi \sin \dfrac {\pi }{3}}=\dfrac {1}{\left( -1\right) \times \cos \dfrac {\pi }{3}-0}=\dfrac {1}{-\cos \dfrac {\pi }{3}}=\dfrac {-1}{\dfrac {1}{2}}=-2 $
