Find the pdf of $\frac{Z_1}{\sqrt{(Z_1^2+Z_2^2)/2}}$ where $Z_1,Z_2$ are i.i.d standard normal Given $Z_1, Z_2$ are i.i.d standard normal random variables.
Let 
$$V:=\frac{Z_1}{\sqrt{(Z_1^2+Z_2^2)/2}}$$
Derive the pdf of $V$.
The numerator and denominator of $V$ are dependent, so the square of it is not F-distribution. 
 A: A possible way to proceed:
Clearly $Z_1^2$ and $Z_2^2$ are independently distributed $\chi^2_1$ random variables.
Then noting that $Z_1=\text{sgn}(Z_1)|Z_1|$, we have $$V=\text{sgn}(Z_1)\sqrt{\frac{2Z_1^2}{Z_1^2+Z_2^2}}=\sqrt{2Z}\text{sgn}(Z_1)$$, where it is a standard exercise to show that $$Z=\frac{Z_1^2}{Z_1^2+Z_2^2}\sim \text{Beta}\left(\frac{1}{2},\frac{1}{2}\right)$$

Thanks to @whuber for mentioning a much easier to understand geometric approach:
Transform to polar coordinates: $$(Z_1,Z_2)\to (R,\Theta)$$ such that $$Z_1=R\cos\Theta\quad,\quad Z_2=R\sin\Theta$$
So our $V$ becomes
$$V=\sqrt{2}\cos\Theta$$
It is easy to verify that $R$ and $\Theta$ are independently distributed. And in particular, $$\Theta\sim U(0,2\pi)$$
This means $\cos\Theta$ has the so called Arcsine distribution with pdf $$f(t)=\frac{\mathbf1_{|t|<1}}{\pi\sqrt{1-t^2}}$$
So we directly get the density of $V$ as 
\begin{align}
f_V(v)&=\frac{1}{\sqrt 2}f\left(\frac{v}{\sqrt 2}\right)
\\\\&=\frac{\mathbf1_{|v|<\sqrt 2}}{\pi\sqrt{2-v^2}}
\end{align}
