Showing that $\sqrt{2+\sqrt3}=\frac{\sqrt2+\sqrt6}{2}$ So I was playing around on the calculator, and it turns out that $$\sqrt{2+\sqrt3}=\frac{\sqrt2+\sqrt6}{2}$$
Does anyone know a process to derive this?

Both sides are clearly positive, and squaring the left hand side shows that
$$\begin{align}  \left(\frac{\sqrt{2}+\sqrt{6}}{2}\right)^2&=\frac{\sqrt{2}^2+2\sqrt{2}\sqrt{6}+\sqrt{6}^2}{2^2}\\&=\frac{2+4\sqrt{3}+6}{4}\\&=2+\sqrt{3}.
\end{align}$$
