The probability density function of the ratio of two normal R.V.s I'm looking for some help with this probability problem.
Here's the question:

Suppose that $X$ and $Y$ are independent standard normal random variables. Show that the probability density function of $Z = X / |Y|$ is given by
  $$
f(t) = \frac{1}{\pi(1+t^2)}, \quad (-\infty < t < \infty).
$$

Thanks for looking through my question.
-updated-
Hi guys, I've looked through the solution and am still thinking, i'm also thinking if its possible to log the expressions and apply convolution theorem... thanks for the answers once again.
-updated-
tried the solution using first method but couldn't figured out how to complete the last step.
Heres how i did it:
$f(x,y) = f_{x}(x)f_{y}(y) = \frac{1}{2\pi}e^{-\frac{1}{2}(x^2 + y^2)}$
let Z = X/|Y|, V = X then x = v, y = v/z
the Jacobian J = $x_{z}y_{v}-x_{v}y_{z} = \frac{v}{z^2}$ 
by Transformation theorem,
$w(z,v)=f(v,\frac{v}{z})|J| = ???$  
Will someone please point out to me how do i proceed from here to obtain f(t)? thanks a lot!
-update 3- this is actually a Cauchy density!! thanks guys, i think i got it figured out.
 A: I imagine you could use the following theorem:

Let $X_1$ and $X_2$ be jointly continuous r.v.'s with joint density $f_{X_1,X_2}$.
Let $Y_1=g(X_1,X_2)$ and $Y_2=g_2(X_1,X_2)$. 

If
1) $y_1=g(x_1,x_2)$ and $y_2=g_2(x_1,x_2)$ can be uniquely solved for $x_1$ and $x_2$ by, say, 
$x_1=h_1(y_1,y_2)$ and $x_2=h_2(y_1,y_2)$
and  
2) $g_1$ and $g_2$ have continuous partials with
$$
J(x_1,x_2) ={\partial g_1\over\partial x_1}{\partial g_2\over\partial x_2} -
{\partial g_1\over\partial x_2}{\partial g_2\over\partial x_1} \ne0,
$$
then
$$
f_{Y_1,Y_2}(y_1,y_2) = f_{X_1,X_2} (x_1,x_2) |J(x_1,x_2)|^{-1},
$$
where
and $x_1=h_1(y_1,y_2)$ and $x_2=h_2(y_1,y_2)$.

Use the above to find the joint distribution of $Y_1=X$ and $Y_2=X/Y$ (with $X_1=X$ and $X_2=Y$); then integrate to obtain the distribution of $Y_2$.

This, by the way, is a problem in Sheldon Ross' ``A first Course in Probability'' (Chapter 6, Theoretical Exercise 33.)
Edit: This may be a ``sledgehammer''. Refer to the post suggested by Sasha...
A: One way is polar coordinates: $(X,Y)= R(\cos \Theta,\sin\Theta)$.  Observe that $X/Y=\cos\Theta/\sin\Theta$.  Find the distribution of $(\cos \Theta,\sin\Theta)$ by thinking about the density of $(X,Y)$.  You'll see that that's easy and the answer you get is simple.  Then think about tangents and cotangents.
A: This is just a variation on the theme of my earlier answer.  Maybe some will find this way of saying it to be clearer.
Write $(X,Y)$ in polar coordinates as $(R,\Theta)$.  Their joint distribution becomes $$\text{constant}\cdot e^{-r^2/2} r\,dr\,d\theta.\tag{1}$$ From $(1)$ we see that $R$, $\Theta$ are independent and $\Theta$ is uniformly distributed in $\mathbb R\bmod 2\pi$.  The ratio $N/N'$ is $\tan\Theta$.  Since $\tan$ has period one-half of a circle, nothing is lost by taking $\Theta$ to be in $(-\pi/2,\pi/2)$.
$$
\frac{d}{d\theta}\Pr(\tan\Theta\le\theta) = \frac{d}{d\theta}\Pr(\Theta\le \arctan\theta) = \frac{d}{d\theta} \frac{\arctan\theta}{\pi} = \frac{1/\pi}{1+\theta^2}.
$$
