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