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  $\dfrac{\sqrt{6}+\sqrt{2}}{2}$

Assume $$\sqrt{a+\sqrt b}=\sqrt c+\sqrt d$$
where $a,b,c,d$ are all rational.
Then
$$a+\sqrt b=c+2\sqrt{cd}+d.$$
By identification
$$\begin{cases}a=c+d,\\b=4cd.\end{cases}$$
From this,
$$4c^2+4cd-4ac=4c^2-4ac+b=0.$$
This is a quadratic equation in $c$, with roots
$$c=\frac{a\pm\sqrt{a^2-b}}2.$$
So for a rational solution to exist, $a^2-b$ must be a perfect rational square.

With $a=2,b=3$, 
$$\sqrt{2+\sqrt3}=\sqrt{\frac12}+\sqrt{\frac32}=\frac{\sqrt2+\sqrt6}2.$$
