How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides I have been asked to prove:

$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$

Which I can easily do by converting the LHS to index form, then squaring it and simplifying it down to get 2, which is equal to the RHS squared, hence proved.
However I know you can't square a side during proof because it generates an extraneous solution. So: how do you go about this proof without squaring both sides? Or can my method be made valid if I do this:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
$$...=...$$
$$2=2$$
$$\lvert\sqrt2\rvert=\lvert\sqrt2\rvert$$
$$\sqrt2=\sqrt2\text{ hence proved.}$$
Cheers in advance :)

You can square both sides in a proof if you note the extraneous solutions are added.
Example $\sqrt{2 + \sqrt{3}} -\sqrt{2 - \sqrt{3}} = k$.
First $2 > \sqrt 3$ so $\sqrt{2 - \sqrt{3}} > 0$ so $k > 0$.  !!!TAKE NOTE OF THAT!!!
$(\sqrt{2 + \sqrt{3}} -\sqrt{2 - \sqrt{3}})^2 = k^2$
$2 + \sqrt3 + 2-\sqrt 3 - 2(\sqrt{2 + \sqrt{3}}\sqrt{2 - \sqrt{3}}) = k^2$
$-2(\sqrt{2 + \sqrt{3}}\sqrt{2 - \sqrt{3}}) = k^2 - 4$
$\sqrt {4 -3} = \frac {4-k^2}{2}$
$4-3 = (\frac {4-k^2}{2})^2$ !!!NOTE!!! $\frac {4-k^2}{2} \ge 0$.
$1 = (\frac {4-k^2}{2})^2$ so
$\frac {4-k^2}{2} = \pm 1$.  !!!BUT we took note that $\frac {4-k^2}{2}\ge 0$.!!!
So $\frac {4-k^2}{2} = 1$
So $4-k^2 = 2$
and $k^2 = 2$ so 
$k = \pm \sqrt 2$ !!!But we took note that $k>0$ so $k = \sqrt 2$.
That is valid.
