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