<s>[INST] 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 :)
 [/INST] This one is a little bit round about.
$\sin \frac {\pi}{12} =  \sin (\frac {\pi}{3} - \frac {\pi}{4})$ by angle addition rules
and
$\sin \frac {\pi}{12} = \sqrt {\frac {1-\cos \frac {\pi}{6}}{2}}$ by the half angle rules.
$\sin (\frac {\pi}{3} - \frac {\pi}{4}) = \frac {\sqrt 6 - \sqrt 2}{4}$
$\sqrt {\frac {1-\cos \frac {\pi}{6}}{2}} = \frac {\sqrt {2-\sqrt 3}}{2}$
$\frac {\sqrt 6 - \sqrt 2}{4} = \frac {\sqrt {2-\sqrt 3}}{2}$
similarly
$\cos \frac {\pi}{12} =  \cos (\frac {\pi}{3} - \frac {\pi}{4}) = \sqrt {\frac {1+\cos \frac {\pi}{6}}{2}}\\
\frac {\sqrt 6 + \sqrt 2}{4} = \frac {\sqrt {2+\sqrt 3}}{2}$
$2\cos \frac{\pi}{12} - 2\sin \frac {\pi}{12} = \frac {\sqrt 6 + \sqrt 2}{2} - \frac {\sqrt 6 - \sqrt 2}{2} = \sqrt 2 = \sqrt {2+\sqrt 3} - \sqrt {2-\sqrt 3}$
</s>