How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$ 
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$

How to change $\sqrt{2+\sqrt{3}}$ into  $\dfrac{\sqrt{6}+\sqrt{2}}{2}$

You can use the identity
\begin{equation}
\sqrt{a+\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} + \sqrt{\frac{a-\sqrt{a^2-b}}{2}}.
\end{equation}
