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