Call the set of points [tex]H[/tex] (for half-disk), where in polar coordinates the region is
[tex]H=\left\{(r,\theta)\mid0<r<1\text{ and }-\dfrac\pi2<\theta<\dfrac\pi2\right\}[/tex]
[tex]H[/tex] has area [tex]\frac\pi2[/tex], and so points in [tex]H[/tex] are uniformly distributed according to the joint density,
[tex]P(R=r,\Theta=\theta)=\begin{cases}\frac2\pi&\text{for }r\in(0,1),\theta\in\left(-\frac\pi2,\frac\pi2\right)\\0&\text{otherwise}\end{cases}[/tex]
If [tex]p(R\cos\Theta,R\sin\Theta)[/tex] is some point in [tex]H[/tex], then the slope of the line through [tex]p[/tex] and the origin is
[tex]S=\dfrac{R\sin\Theta-0}{R\cos\Theta-0}=\tan\Theta[/tex]
Notice that [tex]-\frac\pi2<\Theta<\frac\pi2[/tex] means we can have [tex]-\infty<S<\infty[/tex].
a. The CDF of [tex]S[/tex] is then
[tex]P(S\le s)=P(\tan\Theta\le s)=P(\Theta\le\tan^{-1}s)[/tex]
Compute the marginal density for [tex]\Theta[/tex] by integrating the joint density over the support of [tex]R[/tex]:
[tex]P(\Theta=\theta)=\displaystyle\int_{-\infty}^\infty P(R=r,\Theta=\theta)\,r\,\mathrm dr=\frac2\pi\int_0^1 r\,\mathrm dr[/tex]
[tex]\implies P(\Theta=\theta)=\begin{cases}\frac1\pi&\text{for }\theta\in\left(-\frac\pi2,\frac\pi2\right)\\0&\text{otherwise}\end{cases}[/tex]
Then the marginal CDF for [tex]\Theta[/tex] is
[tex]P(\Theta\le\theta)=\begin{cases}0&\text{for }\theta\le-\frac\pi2\\\frac\theta\pi&\text{for }\theta\in\left(-\frac\pi2,\frac\pi2\right)\\1&\text{for }\theta\ge\frac\pi2\end{cases}[/tex]
It follows that the CDF for [tex]S[/tex] is then
[tex]P(S\le s)=\dfrac\pi2+\dfrac{\tan^{-1}s}\pi[/tex]
where [tex]s\in\Bbb R[/tex].
b. Differentiate the CDF to find the density:
[tex]P(S=s)=\dfrac{\mathrm d}{\mathrm ds}P(S\le s)=\dfrac1{\pi(s^2+1)}[/tex]
