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.

$Z_1 + Z_2, Z_1 - Z_2$ 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 $(1,1)$
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 $\frac{1}{\pi\sqrt{t}(1+t)}$.