Answer
$\dfrac{3}{\sqrt {x^2-9}}$
Work Step by Step
Suppose $\theta =\sin^{-1} (\dfrac{\sqrt {x^2-9}}{x})$
This gives: $\sin \theta=\dfrac{\sqrt {x^2-9}}{x}$
Since, $ r=\sqrt {x^2+y^2}$
We know $ a=\sqrt {x^2-(x^2-9)}=3$
Therefore, we have $\cot [\sin^{-1} (\dfrac{\sqrt {x^2-9}}{x})]=\dfrac{3}{\sqrt {x^2-9}}$
