Covariance in the errors of random variables I have two computed variables say $x\sim N(\mu_{x}, \sigma_{x})$ and $y\sim N(\mu_y, \sigma_y)$. Additionally, the $\sigma_x$ and $\sigma_y$ are both computed from different types of errors (different components used to compute $\mu_x$ and $\mu_y$).
$$\begin{align}
\sigma_x & = \sqrt{A_x^2 + B_x^2 + C_x^2 + D_x^2}\\
\sigma_y & = \sqrt{A_y^2 + B_y^2 + C_y^2 + D_y^2}
\end{align}$$
My goal is to find the covariance in $\sigma_x$ and $\sigma_y$.
I know that (assuming A, B, C, D are independent from each other, thus cross terms are zero) for,
\begin{align}
\text{cov}([A_x, B_x, C_x, D_x], [A_y, B_y, C_y, D_y]) = \text{cov}(A_x, A_y) + \text{cov}(B_x, B_y)+ \text{cov}(C_x, C_y)+ \text{cov}(D_x, D_y)
\end{align}
However, I am stuck when I have to compute $\text{cov}(\sqrt{[A_x^2, B_x^2, C_x^2, D_x^2]}, \sqrt{[A_y^2, B_y^2, C_y^2, D_y^2]})$.
I am not sure if the relation $\sqrt{\text{cov}(A^2, B^2)} = \text{cov}(A, B)$ works.
Any help will be appreciated.
Apologies, if this question is not in the right format to ask.
EDIT:
Following of how $X$ is computed using $A$, $B$, $C$ and $D$,
\begin{align}
X = \dfrac{A}{B} + C + D
\end{align}

Generally speaking, the relation $\sqrt{Cov(A^2,B^2)}=Cov(A,B)$ does not hold. Consider the following counterexample:
Let $X\sim U[0,2\pi]$ and $Y=\sin(x), Z=\cos(X)$. You can see here the proof for $Cov(Y,Z)=0$. Now, let's examine $Cov(Y^2,Z^2)$:
$$Cov(Y^2,Z^2)=E[(Y^2-E[Y^2])(Z^2-E[Z^2])]=E\left[(\sin^2(X)-\int_0^{2\pi}\sin^2(x)dx)(\cos^2(X)-\int_0^{2\pi}\cos^2(x)dx)\right]$$
The result of both integrals is $\pi$ so
$$Cov(Y^2,Z^2)=E\left[(\sin^2(X)-\pi)(\cos^2(X)-\pi)\right]\\=E[\sin^2(X)\cos^2(X)-\pi\sin^2(X)-\pi\cos^2(X)+\pi^2]\\=\int_0^{2\pi}\sin^2(x)\cos^2(x)dx-E[\pi(\sin^2(X)+\cos^2(X))+\pi^2]\\=\int_0^{2\pi}{\sin^2(x)\cos^2(x)dx}-E[\pi\cdot1+\pi^2]\\=\int_0^{2\pi}{\sin^2(x)\cos^2(x)dx}-\pi+\pi^2$$
The result of this integral is $\pi/4$ so overall we get $Cov(Y^2,Z^2)=\pi^2-\frac{3}{4}\pi$ and then $\sqrt{Cov(Y^2,Z^2)}\ne Cov(Y,Z)$.

The covariance of $x,y$ is defined as $Cov(x,y)=E[xy]-E[x]E[y]$. Given $x\sim N(\mu_x,\sigma_x^2), y\sim N(\mu_y,\sigma_y^2)$, we know that $E[x]E[y]=\mu_x \mu_y$ and we're left with finding $E[xy]$.
As explained here, we can write $xy$ as $\frac{1}{4}(x+y)^2-\frac{1}{4}(x-y)^2$ (check it!). For our $x,y$ we get that $$(x+y)\sim N(\mu_x+\mu_y,\sigma_x^2+\sigma_y^2+2\sigma_x\sigma_y),\qquad (x-y)\sim N(\mu_x-\mu_y,\sigma_x^2+\sigma_y^2-2\sigma_x\sigma_y)$$
Denote $S_+=\sqrt{\sigma_x^2+\sigma_y^2+2\sigma_x\sigma_y}$ and $S_-=\sqrt{\sigma_x^2+\sigma_y^2-2\sigma_x\sigma_y}$, we get:
$$\frac{(x+y)}{S_+}\sim N\left(\frac{\mu_x+\mu_y}{S_+},1\right),\qquad \frac{(x-y)}{S_-}\sim N\left(\frac{\mu_x-\mu_y}{S_-},1\right)$$
Next, we can write
$$E[xy]=\frac{1}{4}E[(x-y)^2]-\frac{1}{4}E[(x-y)^2]=\frac{1}{4}E\left[S^2_+\left(\frac{x+y}{S_+}\right)^2\right]-\frac{1}{4}E\left[S^2_-\left(\frac{x-y}{S_-}\right)^2\right]\\=\frac{1}{4}S^2_+E\left[\left(\frac{x+y}{S_+}\right)^2\right]-\frac{1}{4}S^2_-E\left[\left(\frac{x-y}{S_-}\right)^2\right]$$
Now look at $E\left[\left(\frac{x+y}{S_+}\right)^2\right]$: we know that $\frac{(x+y)}{S_+}\sim N\left(\frac{\mu_x+\mu_y}{S_+},1\right)$ so its square has a chi-square distribution with non-centrality parameter $\lambda_+=\frac{(\mu_x+\mu_y)^2}{S_+^2}$, thus $E\left[\left(\frac{x+y}{S_+}\right)^2\right]=1+\lambda_+$. In a similar manner, $E\left[\left(\frac{x-y}{S_-}\right)^2\right]=1+\lambda_-$ where $\lambda_-=\frac{(\mu_x-\mu_y)^2}{S_-^2}$. We overall get:
$$E[xy]=\frac{1}{4}S^2_+E\left[\left(\frac{x+y}{S_+}\right)^2\right]-\frac{1}{4}S^2_-E\left[\left(\frac{x-y}{S_-}\right)^2\right]\\=
\frac{1}{4}(S^2_++(\mu_x+\mu_y)^2)-\frac{1}{4}(S^2_-+(\mu_x-\mu_y)^2)\\=
\frac{1}{4}(\sigma_x^2+\sigma_y^2+2\sigma_x\sigma_y-\sigma_x^2-\sigma_y^2+2\sigma_x\sigma_y+\mu^2_x+2\mu_x\mu_y+\mu^2_y-\mu^2_x+2\mu_x\mu_y-\mu^2_y)\\=\frac{1}{4}(4\sigma_x\sigma_y+4\mu_x\mu_y)$$
And finally
$$Cov(x,y)=E[xy]-E[x]E[y]=\sigma_x\sigma_y$$
