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 :)
 A: 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. 
A: 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: \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}
A: $$a=\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\\a^2=2+\sqrt3 +2-\sqrt3 +2\sqrt{2+\sqrt3}\times\sqrt{2-\sqrt3}\\a^2=2+2-2\sqrt{4-3}\\a^2=2\\a>0\\a=\sqrt2$$
A: This is how I would write it

$$\begin{align}
    \left(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\right)^2
        &= 2 + \sqrt3 + 2 - \sqrt3 - 2 \sqrt{2+\sqrt3}\sqrt{2-\sqrt3}\\
        &= 4 - 2\sqrt{2^2-3}\\
        &= 2
\end{align}$$
  Hence, since $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}⩾0$ ($\sqrt⋅$ being increasing), it follows from the definition of the square root that
  $$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$

The key point here is to remember that the square root of $2$ is by definition¹ the only positive real number $x$ such that $x²=2$.
Also, please don't do the “write equivalent equalities and arrive at something trivially true” thing. Ever. It is never better than directly chaining $=$ and can backfire in interesting ways if one of your $\Leftrightarrow$s is really a $\Rightarrow$.

1. According to my favourite teacher, 99% of maths just follows from definitions, and the other 1% takes your sanity away.
A: $$\left(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\right)^2 \\\\ =\ (2+\sqrt3)+(2-\sqrt3)-2\sqrt{(2+\sqrt3)(2-\sqrt3)} \\\\ =2$$
Also
$$\sqrt{\sqrt2+\sqrt3}-\sqrt{\sqrt2-\sqrt3}>0$$
A: Since $\sqrt{2+\sqrt3}\cdot\sqrt{2-\sqrt3} = 1$ we may write the $\text{LHS}$ as $$\sqrt{2+\sqrt3}\cdot\sqrt{2-\sqrt3}\cdot\left(\sqrt{2+\sqrt 3}-\sqrt{2-\sqrt 3}\right)$$
$$= \left(2+\sqrt 3\right)\sqrt{2-\sqrt3} - \left(2-\sqrt3\right)\sqrt{2+\sqrt3}$$
$$=-2\left(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\right) + \sqrt3\left(\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3}\right)$$
Multiplying and dividing the second term with the original LHS we have 
$$-2x + \sqrt 3\cdot{2+ \sqrt 3 - 2 +\sqrt 3 \over x}$$
$$\implies\frac 6x - 2x = x$$ $$\implies\frac 6x = 3x$$
$$\implies x = \sqrt2$$
A: 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 $a-b=\sqrt{2}\,$.
A: $$$$
\begin{align}
   \sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &= x \tag{A}\\
   \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)\bigg(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\bigg) 
   &= \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x \\
   (2+\sqrt 3)-(2-\sqrt 3) &= \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x \\
   \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x &= 2\sqrt 3 \\
   \sqrt{2+\sqrt3} + \sqrt{2-\sqrt3} &= \dfrac{2\sqrt 3}{x} \tag{B}\\
   2\sqrt{2+\sqrt 3}  &= \dfrac{2\sqrt 3}{x} + x \tag{B+A}\\
   \sqrt{2+\sqrt 3}  &= \dfrac{\sqrt 3}{x} + \dfrac x2 \tag{C}\\
   2\sqrt{2-\sqrt 3}  &= \dfrac{2\sqrt 3}{x} - x \tag{B-A}\\
   \sqrt{2-\sqrt 3}  &= \dfrac{\sqrt 3}{x} - \dfrac x2 \tag{D}\\
   1 &= \dfrac{3}{x^2} - \dfrac{x^2}{4} \tag{CD} \\
   4x^2 &= 12 - x^4 \\
   x^4 + 4x^2 - 12 &= 0 \\
   (x^2 - 2)(x^2 + 6) &= 0 \\
   x &= \sqrt 2 \\
   \sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &= \sqrt 2
\end{align}
A: To be proven:

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

Denote: $a^2=2+\sqrt3,b^2=2-\sqrt3$. Then:
$$a^2+b^2=4\\
a^2b^2=1 \Rightarrow ab=1\\
a^2+b^2+2ab=4+2=6 \Rightarrow a+b=\sqrt6\\
a^2-b^2=2\sqrt3 \Rightarrow (a-b)(a+b)=2\sqrt3 \Rightarrow a-b=\frac{2\sqrt3}{\sqrt6}=\sqrt2.$$
A: 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.
A: Actually, $(\sqrt6\pm\sqrt2)^2=4(2\pm\sqrt3)$ and so $2\sqrt{2\pm\sqrt3}=\sqrt6\pm\sqrt2$. Thus
$$
2(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}) = (\sqrt6+\sqrt2) - (\sqrt6-\sqrt2) = 2\sqrt2
$$
and so
$$
\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} = \sqrt2
$$
