Show that $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$ 
Prove the following:
  $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$

How do you go about proving this?
 A: Hint:
In general  a double radical $\sqrt{a\pm \sqrt{b}}$  can be denested if $a^2-b$ is a perfect square and in this case we have:
$$
\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}
$$
In Your case $a^2-b=1$ so....
A: \begin{align}\sqrt{2-\sqrt{3}} &=\frac{\sqrt{2}\sqrt{2-\sqrt{3}}}{\sqrt{2}}\\
 &=\frac {\sqrt { 4-2\sqrt{3}}}{\sqrt{2}} \\
&=\frac {\sqrt{1-2\sqrt{3} +{ \left( \sqrt{3}\right)}^{2} }  }{ \sqrt {2}  } \\
&=\frac{\sqrt { { \left( 1-\sqrt{3}\right)}^{2}}}{ \sqrt{2}}\\
 &=\frac{\sqrt {3} -1 }{ \sqrt {2}  }\\
& = \frac{\sqrt{6}-\sqrt{2}}{2} \end{align}
A: Previous comments and other answers show you the path to the solution.
I would like to propose another formula of the same kind, thats illustrates the fact that cancellation of nested square roots may sometimes occur ... but not necessarily in any obvious way !
Question Find a simpler form for $x=\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}$
Answer
We have :
$$x\sqrt{2}=\sqrt{4+2\sqrt{3}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(1+\sqrt{3}\right)^{2}}+\sqrt{\left(1-\sqrt{7}\right)^{2}}=1+\sqrt{3}+\sqrt{7}-1=\sqrt{3}+\sqrt{7}$$
Hence : $2x^{2}=10+2\sqrt{21}$ and finally (since $x\ge0$) :
$$\boxed{x=\sqrt{5+\sqrt{21}}}$$
A slightly more general formula could be :
For all $(p,q)\in\mathbb{N}^2$ :
$$\sqrt{p+1+\sqrt{2p+1}}+\sqrt{q+1-\sqrt{2q+1}}=\sqrt{p+q+1+\sqrt{\left(2p+1\right)\left(2q+1\right)}}$$
A: $\sqrt{2-\sqrt3}=\sqrt{\frac{4-2\sqrt3}{2}}=\frac{\sqrt3-1}{\sqrt2}$
