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 :)

\begin{eqnarray}\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &=& \sqrt{4+2\sqrt3 \over 2}-\sqrt{4-2\sqrt3 \over 2}\\ &=&\sqrt{(\sqrt{3} +1)^2 \over 2}-\sqrt{(\sqrt{3} -1)^2\over 2}\\ &=&{\sqrt{3} +1 \over \sqrt{2}}-{\sqrt{3} -1\over \sqrt{2}}\\ &=&\sqrt2 \end{eqnarray}

Let $a=\sqrt{2+\sqrt3}\,$, $b = \sqrt{2-\sqrt3}\,$, then:

$$\require{cancel} a^2+b^2 = 2+\bcancel{\sqrt{3}}+2 - \bcancel{\sqrt{3}} = 4 \\ ab = \sqrt{(2+\sqrt3)(2-\sqrt3)} = \sqrt{2^2 - (\sqrt{3})^2} = \sqrt{4-3} = \sqrt{1} = 1 $$

It follows that: $$(a-b)^2 = a^2+b^2-2ab = 4 - 2 \cdot 1 = 2$$

Since $\sqrt{2+\sqrt3} \gt \sqrt{2-\sqrt3}\,$, $a-b \gt 0$ must be the positive root, so $a-b=\sqrt{2}\,$.

First of all we're not trying to find a solution of the equation here, what you are suggesting is to prove that $\mathrm{lhs} =\sqrt2 $ To do so we square the lhs (first read it fully) and we get $2$. So lhs would be $\sqrt2$ or $-\sqrt2$.

Now we observe the fact that lhs was positive initially ( as $ 2+\sqrt3 > 2-\sqrt3 $) hence lhs would take the positive value ie. $ +\sqrt2$, which is equal to rhs.

So I think it can be solved by observation and easy maths.

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

$\cos \frac {\pi}{12} = \cos (\frac {\pi}{3} - \frac {\pi}{4}) = \sqrt {\frac {1+\cos \frac {\pi}{6}}{2}}\\ \frac {\sqrt 6 + \sqrt 2}{4} = \frac {\sqrt {2+\sqrt 3}}{2}$

$2\cos \frac{\pi}{12} - 2\sin \frac {\pi}{12} = \frac {\sqrt 6 + \sqrt 2}{2} - \frac {\sqrt 6 - \sqrt 2}{2} = \sqrt 2 = \sqrt {2+\sqrt 3} - \sqrt {2-\sqrt 3}$

$$a=\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\\a^2=2+\sqrt3 +2-\sqrt3 +2\sqrt{2+\sqrt3}\times\sqrt{2-\sqrt3}\\a^2=2+2-2\sqrt{4-3}\\a^2=2\\a>0\\a=\sqrt2$$