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.
 A: I think I've found the error in my original method. I'd appreciate feedback on my solution! We will solve this in generality; that is, we will show, as @JimB asserted, that $P(X<kY) = 1/2$ for any k.
The working in my original analysis holds to a point; we do have to find the length of the interval in $[0, 2\pi]$ over which  $f(t) = \cos t - k \sin t < 0$. First, note that $\cos t - k \sin t = 0$ if and only if $\cot t = k$. Because $\cot t$ takes all real values exactly once on each of the intervals $(0, \pi)$ and $(\pi, 2\pi)$, we have exactly two roots of the equation $f(t) = 0$, and they have the form $a$ and $a+\pi$ where $a \in (0, \pi)$.
$f(t) = \cos t - k \sin t$ is continuous and has no other roots, and hence takes either purely negative or purely positive values on the interval $(a, a+\pi)$. But note that $\pi$ is on this interval, and certainly $f(\pi) = -1$ and therefore is negative. Therefore, $f(t)$ is negative on $(a, a+\pi)$.
Also, $f(0) = f(2\pi) = 1$, and once again by continuity and the fact that there are no other roots of $f(t) = 0$, $f(t)$ is positive on the intervals $[0, a)$ and $(a+\pi, 2\pi]$.
Hence, $f(t) = \cos t - k \sin t < 0$ over $(a, a+\pi)$, an interval of length $\pi$. Hence  $P(X<kY) = \frac{\pi}{2\pi} = \frac{1}{2}$.
Edit: Slicker proof. Express $\cos t - k \sin t$ as $\sqrt{1+k^2}(\frac{1}{\sqrt{1+k^2}}\cos t - \frac{k}{\sqrt{1+k^2}}\sin t)= \sqrt{1+k^2}(\cos a\cos t - \sin a \sin t) = \sqrt{1+k^2}(\cos (t+a))$
for some a. This is clearly a scaled and shifted cosine function, and hence is negative over an interval of length $\pi$ for $t \in [0, 2\pi]$.
A: 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:

