Find the Distribution of the following; 
Give that $Z_1,Z_2,...,Z_n$ are independent identically distributed standard Gaussian random variables with mean 0 and variance 1.
  Find the Distribution of $$X=\dfrac{(Z_1+Z_2)^2}{(Z_1-Z_2)^2}$$  

Now first thing I did was open it up, but no common distributions I know are looking familiar. Any hints would be appreciated.
 A: $Z_1 + Z_2, Z_1 - Z_2$ are jointly Gaussian and luckily their convariance is $0$, so they are also independent, then it's easy to see the quotient of their squares follows a F-distribution with parameter $(1,1)$
A: It is useful to recall that, if $Z_1$ and $Z_2$ are independent normal variables with unit variance, then $W=Z_2/Z_1$ has a Cauchy distribution. Obviously $U=\left(\frac{Z_1+Z_2}{Z_1-Z_2}\right)^2=\left(\frac{1+W}{1-W}\right)^2$ is a non-negative random variable, and, for every $t\in[0,1)$:
$$\mathbb{P}[U\leq t]=\mathbb{P}\left[\frac{1+\sqrt{t}}{-1+\sqrt{t}}\leq W\leq\frac{-1+\sqrt{t}}{1+\sqrt{t}}\right]=\frac{2}{\pi}\arctan\sqrt{t}$$
while for $t\in[1,+\infty)$ we have:
$$\mathbb{P}[U\leq t]=\mathbb{P}\left[W\leq\frac{-1+\sqrt{t}}{1+\sqrt{t}}\right]+\mathbb{P}\left[W\geq\frac{1+\sqrt{t}}{-1+\sqrt{t}}\right]=\frac{2}{\pi}\arctan\sqrt{t}$$
hence the pdf of $U$ is supported on $\mathbb{R}^+$ and given by $\frac{1}{\pi\sqrt{t}(1+t)}$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\color{#66f}{\large\pp\pars{X}}&\equiv
\int_{-\infty}^{\infty}\dd Z_{1}\,{\exp\pars{-Z_{1}^{2}/2}\over \root{2\pi}}
\int_{-\infty}^{\infty}\dd Z_{2}\,{\exp\pars{-Z_{2}^{2}/2}\over \root{2\pi}}\,
\delta\pars{X - {\pars{Z_{1} + Z_{2}}^{2} \over \pars{Z_{1} - Z_{2}}^{2}}}
\\[5mm]&={1 \over 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\exp\pars{-\,{Z_{1}^{2} + Z_{2}^{2} \over 2}}
\delta\pars{X - {\bracks{Z_{1} + Z_{2}}^{2} \over \bracks{Z_{1} - Z_{2}}^{2}}}
\,\dd Z_{1}\,\dd Z_{2}
\\[5mm]&={1 \over 2\pi}\int_{0}^{2\pi}\int_{0}^{\infty}
\exp\pars{-\,{\rho^{2} \over 2}}
\delta\pars{X - {\bracks{\cos\pars{\theta} + \sin\pars{\theta}}^{2} \over \bracks{\cos\pars{\theta} - \sin\pars{\theta}}^{2}}}\rho\,\dd\rho\,\dd\theta
\\[5mm]&=\color{#66f}{\large{\Theta\pars{X} \over 2\pi}\int_{-\pi}^{\pi}
\delta\pars{%
X - {\bracks{\cos\pars{\theta} + \sin\pars{\theta}}^{2} \over \bracks{\cos\pars{\theta} - \sin\pars{\theta}}^{2}}}
\,\dd\theta}
\\[5mm]&\mbox{because}\quad
\int_{0}^{\infty}\exp\pars{-\,{\rho^{2} \over 2}}\rho\,\dd\rho = 1
\end{align}

Also,
  \begin{align}
&\color{#66f}{\large\pp\pars{X}}={\Theta\pars{x} \over 2\pi}\int_{-\pi}^{\pi}
\delta\pars{%
X - {\cos^{2}\pars{\theta - \pi/4} \over \cos^{2}\pars{\theta + \pi/4}}}\,\dd\theta
\\[5mm]&={\Theta\pars{x} \over 2\pi}\bracks{\int_{0}^{\pi}\delta\pars{%
X - {\cos^{2}\pars{\theta - \pi/4}\over\cos^{2}\pars{\theta + \pi/4}}}\,\dd\theta
+\int_{0}^{\pi}\delta\pars{%
X - {\cos^{2}\pars{\theta + \pi/4}\over\cos^{2}\pars{\theta - \pi/4}}}\,\dd\theta}
\\[5mm]&={\Theta\pars{x} \over 2\pi}\bracks{\int_{-\pi/2}^{\pi/2}\!\delta\pars{%
X - {\sin^{2}\pars{\theta - \pi/4}\over\sin^{2}\pars{\theta + \pi/4}}}\,\dd\theta
+\int_{-\pi/2}^{\pi/2}\!\delta\pars{%
X - {\sin^{2}\pars{\theta + \pi/4}\over\sin^{2}\pars{\theta - \pi/4}}}\,\dd\theta}
\\[5mm]&={\Theta\pars{x} \over \pi}\bracks{\int_{0}^{\pi/2}\!\delta\pars{%
X - {\sin^{2}\pars{\theta - \pi/4}\over\sin^{2}\pars{\theta + \pi/4}}}\,\dd\theta
+\int_{0}^{\pi/2}\!\delta\pars{%
X - {\sin^{2}\pars{\theta + \pi/4}\over\sin^{2}\pars{\theta - \pi/4}}}\,\dd\theta}
\\[5mm]&={\Theta\pars{x} \over \pi}\bracks{\int_{-\pi/4}^{\pi/4}\!\delta\pars{%
X - {\sin^{2}\pars{\theta}\over\cos^{2}\pars{\theta}}}\,\dd\theta
+\int_{-\pi/4}^{\pi/4}\!\delta\pars{%
X - {\cos^{2}\pars{\theta}\over\sin^{2}\pars{\theta}}}\,\dd\theta}
\\[5mm]&={2\Theta\pars{x} \over \pi}\bracks{\int_{0}^{\pi/4}\!\delta\pars{%
X - \tan^{2}\pars{\theta}}\,\dd\theta
+\int_{0}^{\pi/4}\!\delta\pars{X - \cot^{2}\pars{\theta}}\,\dd\theta}
\\[5mm]&={2\Theta\pars{x} \over \pi}\left\{\Theta\pars{1 - X}\int_{0}^{\pi/4}{\delta\pars{\theta - \arctan\pars{\root{X}}} \over
\verts{2\tan\pars{\theta}\sec^{2}\pars{\theta}}}\,\dd\theta\right.
\\&\left.\phantom{{2\Theta\pars{x} \over \pi}\left[\right.}
+\Theta\pars{X - 1}\int_{0}^{\pi/4}
{\delta\pars{\theta - {\rm arccot}\pars{\root{X}}} \over
\verts{2\cot\pars{\theta}\bracks{-\csc^{2}\pars{\theta}}}}\,\dd\theta\right\}
\\[5mm]&={2\Theta\pars{x} \over \pi}\bracks{%
{\Theta\pars{1 - X} \over 2\root{X}\pars{X + 1}}
+{\Theta\pars{X - 1} \over 2\root{X}\pars{X + 1}}}
=\color{#66f}{\large{\Theta\pars{X} \over \pi}\,{1 \over \root{X}\pars{X + 1}}}
\end{align}

