Is $X^2-Y^2/ \sqrt{X^2+Y^2}$ normal where $X,Y\sim N(0,1)$ Can I say that  $(X^2-Y^2)/ \sqrt{X^2+Y^2}$ is normal since $X,Y\sim N(0,1)$ and $X,Y$ are normal and independent?
I am trying to do those problem using polar coordinates with $X=r \cos \theta$ and $Y=r \sin \theta$, but I got stuck when I was trying to simplify $(X^2-Y^2)/ \sqrt{X^2+Y^2}$. Can anybody please help me? Thanks!

Let $(R,\Theta)$ be the polar coordinates corresponding to the Cartesian coordinates $(X,Y)$, with, say, $0 \le \Theta < 2 \pi$. Thus $X = R \cos(\Theta)$, $Y = R \sin(\Theta)$, 
$R = \sqrt{X^2 + Y^2}$.  Then $R$ and $\Theta$ are independent, and $\Theta$ is
uniform on $[0,2\pi)$.
Now $(X^2 - Y^2)/\sqrt{X^2 + Y^2} = (R^2 \cos^2(\Theta) - R^2 \sin^2(\Theta))/R
 = R \cos(2\Theta)$. But $\cos(2\Theta)$ has the same distribution as $\cos(\Theta)$ (i.e. $2 \Theta \mod 2 \pi$ is again
uniform on $[0,2\pi)$) and is independent of $R$.  So $R \cos(2\Theta)$ has the
same distribution as $R \cos(\Theta) = X$.
