problem involving a bivariate gaussian I need some help with this exercise, I tried to do this but my calculations seem to go nowhere, any help or hint can be very useful

 A: $\newcommand{\var}{\operatorname{var}} \newcommand{\cov}{\operatorname{cov}}$
To be independent they must be uncorrelated, i.e. the covariance must be $0$.  I'm going to use capital $X,Y,U,V$ to refer to the random variables.  We have
$$
\var(X) = \sigma_X^2, \quad\var(Y) = \sigma_Y^2, \quad \cov(X,Y) = \rho\sigma_X\sigma_Y.
$$
Then
$$
\begin{align}
& \cov(U,V) \\[8pt] ={} & \cov(\ (\cos\theta) X -(\sin\theta)Y,\  (\sin\theta)X+(\cos\theta)Y\  ) \\[8pt]
= {} & (\cos\theta)\cov(\  X,\ (\sin\theta)X+(\cos\theta)Y\  ) - (\sin\theta)\cov(\  Y,\   (\sin\theta)X + (\cos\theta)Y\  ) \\[8pt]
= {} & (\cos\theta\sin\theta)\var(X) +(\cos^2\theta)\cov(X,Y)-(\sin^2\theta)\cov(Y,X) - (\sin\theta\cos\theta)\var(Y) \\[8pt]
= &  (\cos\theta\sin\theta)(\sigma^2_X-\sigma^2_Y) + (\cos^2\theta-\sin^2\theta)\rho\sigma_X\sigma_Y.
\end{align}
$$
What must $\theta$ be in order that this last quantity be $0$?
Recall that $2\sin\theta\cos\theta=\sin(2\theta)$ and $\cos^2\theta-\sin^2\theta=\cos(2\theta)$.  So we need
$$
\frac 1 2 \sin(2\theta) (\sigma^2_X-\sigma^2_Y) + (\cos(2\theta)) \rho\sigma_X \sigma_Y = 0.
$$
$$
\sin(2\theta) (\sigma^2_X-\sigma^2_Y) = -2 (\cos(2\theta)) \rho\sigma_X \sigma_Y
$$
$$
\tan(2\theta) = \frac{2\sigma_X \sigma_Y}{\sigma_Y^2 - \sigma_X^2}
$$
$$
\frac{2\tan\theta}{1-\tan^2\theta} =\tan(2\theta) = \frac{2\sigma_X \sigma_Y}{\sigma_Y^2 - \sigma_X^2} = \frac{2\left(\frac{\sigma_X}{\sigma_Y}\right)}{1 - \left(\frac{\sigma_X}{\sigma_Y}\right)^2}
$$
So we need $\tan\theta=\dfrac{\sigma_X}{\sigma_Y}$, so $\theta=\arctan\dfrac{\sigma_X}{\sigma_Y}$.
