## Thomas' Calculus 13th Edition

$x=\cot^2 \theta$;$y= \cot \theta$; $0\leq \theta \leq \dfrac{\pi}{2}$
Let us consider that $y=\sqrt x$ Since, $y=x \tan \theta$ Then, we have $\sqrt x= \cot \theta \implies x=\cot^2 \theta$ Now, $y=(\cot^2 \theta) (\tan \theta)$ Thus, $y=\dfrac{1}{\tan \theta}(\tan \theta)= \cot \theta$