<s>[INST] Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$ It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$
Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}{4})$
I was thinking of making polar transformations $X=r \cos \theta, Y=r \sin \theta$
Then I am getting stuck at ranges of $\theta$
 [/INST] If you have already proved $\frac{X Y}{\sqrt{X^2 + Y^2}} $ and $\frac{X^2 - Y^2}{\sqrt{X^2 + Y^2}}$ are gaussian, as long as that pair of them is jointly gaussian, then you may use their property: $u,\, v \text{ independent} \iff \operatorname{cov}(u,v) = 0$.
$$ \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\frac{X^2 - Y^2}{\sqrt{X^2 + Y^2}} \right)$$
$$ = \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\frac{X^2}{\sqrt{X^2 + Y^2}} \right) - \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\frac{Y^2}{\sqrt{X^2 + Y^2}} \right)$$
Now make use of the symmetry of the expressions we have:
$$ \operatorname{cov}(\cdots) = 0 $$
To make this more rigorous one may rewrite the covariance as follows:
$$ \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\frac{X^2}{\sqrt{X^2 + Y^2}} \right) - \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\frac{Y^2}{\sqrt{X^2 + Y^2}} \right) $$
$$ = E \left( \frac{X^3 Y}{X^2 + Y^2} \right) - E \left( \frac{X Y^3}{X^2 + Y^2} \right) $$ 
Now renaming $X \to Y, \, Y \to X$ under the first expectation sign (note that this is the same as renaming variables under integral) we get the result. 
Covariance method can be carried even further, following the last line and the trick $ X = -X $ ($X$ is symmetric distribution) which we apply to the second $E$:
$$ = E \left( \frac{X^3 Y}{X^2 + Y^2} \right) - E \left( \frac{X Y^3}{X^2 + Y^2} \right) = E \left( \frac{X^3 Y}{X^2 + Y^2} \right) + E \left( \frac{X Y^3}{X^2 + Y^2} \right) $$
$$ = E \left( \frac{X Y(X^2 + Y^2)}{X^2 + Y^2} \right)  = E(XY) = 0$$
The last equation is due to independence of $X, Y$ $\implies 0=\operatorname{cov}(X,Y)=E(XY) - E(X)E(Y)=E(XY).$
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