\(\dfrac{13\pi}6\equiv\dfrac\pi6\mod2\pi\) (in other words, they represent the same angle on the unit circle)
So,
\(\sec\dfrac{13\pi}6=\sec\dfrac\pi6=\dfrac1{\cos\dfrac\pi6}\)
You have \(\cos\dfrac\pi6=\dfrac{\sqrt3}2\), so \(\sec\dfrac\pi6=\dfrac{\sqrt3}2\)
