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?

The simpliest way is to square both sides and then compare them.
Formally, consider $\left(\frac{\sqrt{2}+\sqrt{6}}{2}\right)^{2}=\left(\frac{8+2\sqrt{12}}{4}\right)=2+\sqrt{3}$.
Taking square-root on both sides yields $\frac{\sqrt{2}+\sqrt{6}}{2}=\sqrt{2+\sqrt{3}}$.
