$X_{1},X_{2} \sim N(0,1)$ and are independent. Show $\frac{X_{1}}{X_{2}}$ and $\sqrt{X_{1}^{2}+X_{2}^{2}}$ are independent I know $U=\frac{X_{1}}{X_{2}}$ is a Cauchy distribution and $V=\sqrt{X_{1}^{2}+X_{2}^{2}}$ is a Rayleigh distribution since $X_{1},X_{2} \; iid\sim N(0,1)$.
I thought that I can find r.v. $R$ and $\theta$ so that $X_{1}=Rsin\theta$ and $X_{2}=Rcos\theta$. So, $P(U|V)=P(U=tan\theta|V=R)=P(U=tan\theta)=P(U) \Rightarrow U,V\;independent$.
But... do there really exist such r.v. $R$ and $\theta$ $s.t. X_{1}=Rsin\theta,X_{2}=Rcos\theta\;iid \sim N(0,1)$? 
And is my idea correct to solve the problem?
 A: Since $X_1,X_2$ are independent, the joint PDF is 
$$\tag{1}f_{X_1,X_2}(x_1,x_2) = \frac{1}{\sqrt{2\pi}}e^{-x_1^2/2}\frac{1}{\sqrt{2\pi}}e^{-x_2^2/2} = \frac{1}{2\pi}e^{-(x_1^2 + x_2^2)/2}$$
For $(X_1,X_2) \in \mathbb{R}^2$ we have $(U,V) \in (-\infty, \infty)\times [0,\infty)$ under the mapping  
$$U = \frac{X_1}{X_2}, \quad V = \sqrt{X_1^2 + X_2^2}$$
However, it is not one-to-one since points $(X_1,X_2)$ and $(-X_1,-X_2)$ have the same image. There is also a problem of how $U$ is defined when $X_2 = 0$, but this can be ignored by setting $U$ to be $0$ on this set of probability mass $0$.
The joint PDF of $U,V$ can be obtained as 
$$\tag{2}f_{U,V}(u,v) = 2f_{X_1,X_2}(x_1(u,v),x_2(u,v)) |J(u,v)|,$$
where the factor of $2$  accounts for the two-to-one nature of the mapping and the Jacobian $J$ is obtained from the inverse mapping $x_1 = \frac{uv}{\sqrt{1+u^2}}, x_2 =\frac{v}{\sqrt{1 + u^2}} $ as 
$$\tag{3}|J(u,v)| = \left|\frac{\partial(x_1,x_2)}{\partial(u,v)} \right|= \frac{\partial x_1}{\partial u }\frac{\partial x_2}{\partial v }-\frac{\partial x_1}{\partial v }\frac{\partial x_2}{\partial u } = \frac{v}{1 + u^2}$$
Substituting into (2) using (1) and (3) we obtain
$$f_{U,V}(u,v) = \frac1{\pi(1+u^2)}ve^{-v^2/2}$$ 
Since the joint PDF factors into a product of a function of $u$ and a function of $v$, the random variables $U$ and $V$ are independent.
A: Just An idea:
Fact 1: A random vector $x\sim N(0, I_n)$ has the property that
$\|x\|$ and $\frac{x}{\|x\|}$ are independent where $\|\cdot \|$
is the Euclidean norm.
Fact 2: Functions of independent random variables are also independent.
From Facts 1 and 2, the desired result follows. Indeed, for $n=2$,
$\frac{x}{\|x\|} = [\frac{x_1}{\sqrt{x_1^2+x_2^2}}, \ \frac{x_2}{\sqrt{x_1^2+x_2^2}}]$
and $\|x\| = \sqrt{x_1^2 + x_2^2}$ are independent, 
so are $\frac{x_1}{x_2}$ and $\sqrt{x_1^2 + x_2^2}$.
