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?

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....
