= Tangent of a uniform angle has a Cauchy distribution
{title2=$s\tan\Theta\sim\operatorname{Cauchy}(0,s)$}
If $s>0$ and $\Theta$ has the <continuous uniform distribution> on $(-\pi/2,\pi/2)$, then $Y=s\tan\Theta$ has the <Cauchy distribution> with location zero and scale $s$. The <tangent function> is strictly increasing on this interval, so its <cumulative distribution function> is
$$
P(Y\le y)=\frac12+\frac1\pi\arctan\frac ys.
$$
Differentiating gives the <probability density function> $s/[\pi(s^2+y^2)]$. The <probability> of either endpoint is zero, so including endpoints in the uniform interval does not alter this <probability distribution>. The slow quadratic decay makes the <expected value> undefined: its positive and negative parts both have infinite integrals. Its second <moment> is also infinite, so it has no finite <standard deviation>.
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