Ratio of two independent normal : cumulative sum 
Given two independent normal variables, $X\sim N(0, \sigma^2_X)$ and $Y\sim N (0, \sigma_Y^2)$, find the probability that $\frac{X}{Y} < 1$.

I seem to prove: $Z = \frac X Y \sim $ Cauchy distribution with pdf $f_Z(z) = \frac{1}{\pi} \frac{\frac{\sigma_X}{\sigma_Y}}{z^2 + \frac{\sigma^2_X}{\sigma^2_Y}}$ (I'm not sure though). 
Anyway, help would be appreciated.
EDIT Note that the ratio follows Cauchy distribution is something I stumbled on while trying find the probability. As aleady noted on comments, this is not mandatory, and can be done by other technique. The linked question only wants to prove the distribution is Cauchy necessarily. 
 A: Let's generalize the problem in order to draw out the concepts on which it rests.
To this end, let $(U,V)$ be any bivariate random variable whose distribution is invariant under all rotations around the origin. Let $p$ be the probability of the punctured line $U\ne0\, V=0.$ The event $(U,V)\ne 0$ is the disjoint union of infinitely many such punctured lines, each with probability $p$ by the rotational invariance. Since the chance of this union is finite, it must be that $p=0,$ implying the distribution of these lines is continuous.  The angle made by such a line is a value in $[0,\pi)$ given by the angle between the positive $u$ axis and any point on the line in the upper half plane).  The rotational invariance implies this angle has a uniform distribution.  Because the chance $(U,V)$ lies on one of these lines is $1-\Pr(0,0),$ the density of this uniform distribution must be $(1-\Pr(0,0))/\pi.$
Let  $\sigma_X$ and $\sigma_Y$ be positive numbers and $(X,Y) = (\sigma_X U, \sigma_Y V).$  We seek a formula for
$$\Pr\left(\frac{X}{Y} \le a\right)$$
with $a=1$ (generalizing to any $a\gt 0$).
Rewriting this event as
$$a \ge \frac{X}{Y} = \frac{\sigma_X U}{\sigma_Y V} = \frac{\sigma_X}{\sigma_Y} \frac{U}{V}$$
reduces the problem to finding
$$\Pr\left(\frac{U}{V} \le \frac{a\,\sigma_Y}{\sigma_X}\right).$$
This event is the union of (a) the origin $(0,0)$ and (b) all punctured lines making angles between $\operatorname{arccot}(a\sigma_Y/\sigma_X)$ and $\pi.$  Computing the probability in (b) according to the uniform distribution previously found and adding in the chance of (a) gives

$$\Pr\left(\frac{X}{Y} \le a\right) = \Pr(0,0) + \frac{1-\Pr(0,0)}{\pi}\left(\pi - \operatorname{arccot}(a\,\sigma_Y/\sigma_X) \right).$$

In the question $(X,Y)$ has a continuous distribution, whence $\Pr(0,0)=0.$ The result simplifies to

$$\Pr\left(\frac{X}{Y} \le a\right) = 1 -  \frac{\arctan\left(\sigma_X/\left(a\,\sigma_Y\right)\right)}{\pi}.$$

