The Problem: Let $(X,Y)$ be a uniformly chosen random point on the unit circle. Show that $Z=X/Y$ is absolutely continuous and find its PDF.
My Thoughts: To try to compute the CDF of $Z$, I started with $F_Z(t)=P\left(\frac{X}{Y}\leq t\right)=P(x\leq Yt)$ , but did not see a way to proceed from here. I am having issues with visualizing the geometry of the problem and really understanding how to bring the CDF of the uniformly chosen point into play.
Could anyone please provide a hint to get me started on this problem?
Thank you for your time and highly appreciate any help and feedback.
Edit: As suggested by Marc in his insightful comment, we can model the experiment by letting $X=\cos\theta$ and $Y=\sin\theta$ where $\theta\thicksim\text{Unif}[0,2\pi)$. Then I proceed with
\begin{equation}\begin{split}
P(Z\leq t)&=P(\cot\theta\leq t)=P\left(\theta\in\bigcup_{k\in\mathbb Z}(\cot^{-1}(t)+k\pi,(k+1)\pi)\right)\\
&=\sum_{k\in\mathbb Z}F_\theta((k+1)\pi)-F_\theta(\cot^{-1}(t)+k\pi).\end{split}\end{equation}
Then we can differentiate the sum above since all but finitely many terms are nonzero, as noted by Marc in another helpful comment.
Best Answer
The key point as noted by Marc is to use the parametrization $(X,Y)=(\cos(\Theta),\sin(\Theta))$ and $\Theta$ is then uniformly distributed on $[0,2\pi]$.
Noting that $(X,Y)$ has the same distribution as $(Y,X)$ gives a small shortcut since, for $t \in \mathbb R$,
$$P(X/Y \le t) = P(Y/X \le t)= P(\tan(\Theta) \le t)$$
Then the computation is straightforward (using e.g. periodicity):
$$P(\tan(\Theta) \le t) = \pi/2 + Arctan(t)/\pi = \int_{-\infty}^t \frac{1}{ \pi} \frac{1}{1+x^2} dx$$
at which point you shall (perhaps) recognize the Cauchy distribution.