## Precalculus (6th Edition) Blitzer

$\dfrac{\sqrt 2}{x}$
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 $\cot [\tan^{-1} (\dfrac{x}{\sqrt 2})]=\dfrac{\sqrt 2}{x}$