<s>[INST] 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.
 [/INST] Since $$\sqrt{2+\sqrt{3}} = \sqrt{4+2\sqrt{3}\over 2} = {1+\sqrt{3}\over \sqrt{2}}$$
\begin{eqnarray}\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}&=& \frac{\sqrt{2}+{1+\sqrt{3}\over \sqrt{2}}}{\sqrt{2}-{1+\sqrt{3}\over \sqrt{2}}}\\
&=& \frac{2+1+\sqrt{3}}{2-1-\sqrt{3}}\\
&=& \frac{3+\sqrt{3}}{1-\sqrt{3}}=...\\
&=&-3-2\sqrt{3}
\end{eqnarray}
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