Answer
$\dfrac{\sqrt {x^2+4}}{2}$
Work Step by Step
Suppose $\theta =\sin^{-1} (\dfrac{x}{\sqrt {x^2+4}})$
This gives: $\sin \theta=\dfrac{x}{\sqrt {x^2+4}}$
Since, $ r=\sqrt {x^2+y^2}$
We know $ a=\sqrt {x^2+4-x^2}=2$
Therefore, we have $\sec [\sin^{-1} (\dfrac{x}{\sqrt {x^2+4}})]=\dfrac{\sqrt {x^2+4}}{2}$
