Answer
$\frac{1}{u}$
Work Step by Step
1. Let $cot^{-1}u=t, 0\lt t\lt\pi$, we have $cot(t)=u$.
2. Let $x=u,y=1$, we have $r=\sqrt {u^2+1}$
3. Thus $tan(cot^{-1}u)=tan(t)=\frac{y}{x}=\frac{1}{u}$
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