Let $a_0 = \sqrt2 + \sqrt 3 + \sqrt 6$ and let $a_{n+1}=\frac{(a_n)^2-5}{2(a_n)+2}$ for $n\ge0$. Prove that $a_n=\cot\left(\frac{2^{n-3}\pi}{3}\right)-2$ for all n.
I have already proved that $a_0 = \sqrt2 + \sqrt 3 + \sqrt 6$ works for $a_n=\cot\left(\frac{2^{n-3}\pi}{3}\right)-2$ when $n=0$. However, I do not know what to do next. I have tried many things that seem to lead me nowhere. Can someone help point me into the right direction?
Hint. One may recall that $$ \tan(2x)=\frac{2\tan x}{1-\tan^2 x} $$ which reads $$ \cot(2x)=\frac{\cot^2 x-1}{2\cot x} $$ then put $x:=\dfrac{2^{n-3}\pi}{3}$ to get a link between $a_{n+1}$ and $a_n$.