Let $X$ and $Y$ be independent standard normal random variables and $p\in (0,1)$. Show that $\mathbf P(\frac{Y^2}{X^2+Y^2}\leq p)=2/\pi\arcsin\sqrt{p}$.
How to prove this? Apparently the random variable $\frac{Y^2}{X^2+Y^2}$ seems close to the square of a t-distribution with 2 degrees of freedom, but this consideration didn't bring me very far.
Rearrange the above to be $$ \mathbb{P}\left(\frac{Y^2}{X^2} < \frac{p}{1 - p}\right) \Leftrightarrow \mathbb{P}\left(-\sqrt{\frac{p}{1 - p}} < \frac{Y}{X} < \sqrt{\frac{p}{1 - p}}\right).$$ Then use the property that if $Y$ is normally distributed and $X^2$ is chi-squared (with 1 degree of freedom here) distributed then $X/Y$ is a Student-t distribution with $\nu = 1$ degrees of freedom (in this case).
For $\nu = 1$, the probability density function of a Student-t distribution is $f(t) = \frac{1}{\pi(1 + t^2)}$. From the above form of your question and setting $\alpha = \sqrt{p/(1 - p)}$ you get that the answer is equal to: $$ \int_{-\alpha}^{\alpha} f(t) dt = \frac{1}{\pi} [\arctan(\alpha) - \arctan(-\alpha)] = \frac{2}{\pi} \arctan{a} = \frac{2}{\pi} \arcsin{\frac{\alpha}{\sqrt{\alpha^2 + 1}}}.$$
The equalities follow from standard properties of the arctan function.