Answer
$\dfrac{2}{\sqrt {3}}$ or, $\dfrac{2\sqrt {3}}{3}$
Work Step by Step
Suppose $\theta =\sin^{-1} (-\dfrac{1}{2})$
This gives: $\sin \theta=-\dfrac{1}{2}$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ x=\sqrt {r^2-y^2}=\sqrt {2^2-(1)^2}=\sqrt 3$
Therefore, we have $\sec \theta =\sec [\sin^{-1} (-\dfrac{1}{2})]=\dfrac{2}{\sqrt {3}}=\dfrac{2\sqrt {3}}{3}$
