Show that $\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$ 
Show that $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}.$$

My attempt: I could find a way to develop the LHS by un-nesting the double radical, figuring out that
$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}\ \ (1)$$ 
By substituting (1) in the LHS of the original expression, 
it is straightforward to show that it is equal to the RHS. 
But my problem is on how to show without the un-nesting, by another approach. I've used a standard approach of
multiplying both terms by $\sqrt{2}+\sqrt{2+\sqrt{3}}$, leading to 
$$\frac{4+\sqrt{3}+2\sqrt{4+2\sqrt{3}}}{-\sqrt{3}}\Leftrightarrow \frac{(4+\sqrt{3}+2\sqrt{4+2\sqrt{3}})\sqrt{3}}{-3}.$$
But I was not able to show that this last expression is equal to RHS by this approach. 
Hints and answers not using my first un-nesting approach (if possible) will be appreciated. Sorry if this is a duplicate.

we have to show,
$$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$$
Let,$2+\sqrt{3}=p$
$$R.H.S.= -3-2\sqrt{3}$$
$$=-2(\frac{3}{2}+\sqrt{3})$$
$$=-2\bigg(\frac{3}{2}+\frac{1}{2}-\frac{1}{2}+\sqrt{3}\bigg)$$
$$=-2(2+\sqrt{3}-\frac{1}{2})$$
$$=-2(p-\frac{1}{2})$$
$$=1-2p$$
$$=\dfrac{(2-p)(1-2p)}{2-p}$$
$$=\dfrac{2-5p+2p^2}{2-p}...........(1)$$
Now,
$$L.H.S=\dfrac{\sqrt{2}+\sqrt{p}}{\sqrt{2}-\sqrt{p}}$$
$$=\dfrac{(\sqrt{2}+\sqrt{p})^2}{2-p}.............(2)$$
Now,we can conclude that if the given statement is true then,for $p=2+\sqrt{3}$,the numerators of part (1) and (2) will be equal.in reverse we can also say that the numerators will be same if $p=2+\sqrt{3}$.So,now my target is to solve for $p$ to get $2+\sqrt{3}$.
to be the statement true,
$$(\sqrt{2}+\sqrt{p})^2=2-5p+2p^2$$
$$p^2-3p-\sqrt{2}p=0$$
$$p=0,2+\sqrt{3}$$
As  we got a root $p=2+\sqrt{3}$,then the claim is true.$_{[Showed]}$
NOTE: you can also prove it by substituting the value of p in (2) and getting (1).but I think the above procedure is more simple.
