Obtaining a tractable expression for P(X<2Y) The joint probability of the bivariate random vector $(X,Y)$ is given by $f_{X, Y}(x,y) = \frac{1}{2\pi}e^{-\sqrt{x^2+y^2}}$. Compute $P(X<2Y)$.
My attempt: I made the obvious transformation to polar coordinates to obtain $f_{R, \theta}(r,t) = \frac{1}{2\pi}re^{-r}$. This is pleasant; $R$ and $\theta$ are independent of each other, with $\theta$ being uniform on $[0, 2\pi]$ and $R$ being a gamma. Since we have defined $X = R \cos\theta$ and $Y = R \sin\theta, X<2Y$ whenever $\cos\theta<2\sin\theta$, or $\cot\theta<2$ (edit: this is a blunder, see below). The uniformity of $\theta$ over $[0, 2\pi]$ means that $P(X<2Y)$ is the total length of the intervals on which $\cot\theta<2$, divided by $2\pi$.
...is that as far as a straightforward computation goes? Maybe I'm missing something obvious, but I know no way of numerically evaluating what I found here. Is there a better way to solve this, which obtains a more tractable expression for the probability?
Edit: there is a blunder in my analysis: $\cos\theta<2\sin\theta$ does not imply $\cot\theta<2$! I got sloppy with my signs.

This answer is somewhat in the spirit of the video listed in https://mathematica.stackexchange.com/questions/267065/how-to-visualize-the-circle-triangle-probability-problem.
And it is a bit of a handwaving proof based on the fact that the bivariate pdf is radially symmetric about $(0,0)$.  So no matter what value of $k$ is used for determining $Pr(X < k Y)$ the result is always 1/2 as the plane defined by $X = k Y$ divides the pdf into two equal parts:

