Why is $5\tan(54^\circ) = \sqrt{25 + 10\sqrt{5}}$ and $\tan\left(\frac{\pi}{5}\right) = \sqrt{5 - 2\sqrt{5}}$? On the Wikipedia Page about Pentagons, I noticed a statement in their work saying that $\sqrt{25+10\sqrt{5}}=5\tan(54^{\circ})$ and $\sqrt{5-2\sqrt{5}}=\tan(\frac {\pi}{5})$
My question is: How would you justify that? My goal is to simplify $\sqrt{25+10\sqrt{5}}$ and $\sqrt{5-2\sqrt{5}}$ without knowledge on the detested radical!

As $54\cdot5\equiv90\pmod{180},$
if $\tan5x=\infty,5x=180^\circ n+90^\circ$ where $n$ is any integer
$x=36^\circ n+18^\circ$ where $n\equiv0,\pm1,\pm2\pmod5$
Using $\tan5x$ expansion, the roots of $$5t^4-10t^2+1=0$$ are $\tan(36^\circ n+18^\circ)$  where $n\equiv0,\pm1,2\pmod5$ 
as $n=2\implies\tan(36^\circ n+18^\circ)=?$
$$t^2=\dfrac{10\pm4\sqrt5}{2\cdot5}=\dfrac{5\pm2\sqrt5}5$$
As $\tan54^\circ>\tan18^\circ>0,\tan^254^\circ>\tan^218^\circ,$
$\tan^254^\circ=\dfrac{5+2\sqrt5}5=\dfrac{25+10\sqrt5}{5^2}$
and $\tan^218^\circ=\dfrac{25-10\sqrt5}{5^2}$
For $\tan36^\circ,$ can you use the same idea?
