Prove $\sum_{n=1}^\infty \text{arccot }a_n^2=\frac{\pi}{12}$ where $a_n=\frac{\left(2+\sqrt{3}\right)^n-\left(2-\sqrt{3}\right)^n}{\sqrt{3}}$

Let $a_1=2$, $a_2=8$, $a_n=4a_{n-1}-a_{n-2}$, $n=3,4,5,\ldots$. Prove: $$\sum_{n=1}^\infty \mathrm{arccot} (a_n^2)=\frac{\pi}{12} $$

My attempt: I have worked out $a_n=\frac{\left(2+\sqrt{3}\right)^n-\left(2-\sqrt{3}\right)^n}{\sqrt{3}}$, but I do not know how to do it afterwards. Can anyone give me some suggestions? Thank you.

First we have $$a^2_{n}-a_{n+1}a_{n-1}=4$$ Proof: $$a_{n}(4a_{n-1})=a_{n-1}(4a_{n})$$ then we $$a_{n}(a_{n}+a_{n-2})=a_{n-1}(a_{n+1}+a_{n-1})$$ then $$a^2_{n}-a_{n+1}a_{n-1}=a^2_{n-1}-a_{n}a_{n-2}=\cdots=a^2_{2}-a_{3}a_{1}=4$$

so $$\cot^{-1}{a^2_{n}}=\cot^{-1}{\dfrac{a_{n}(4a_{n})}{4}}=\cot^{-1}{\dfrac{a_{n}(a_{n+1}+a_{n-1})}{a^2_{n}-a_{n+1}a_{n-1}}}=\cot^{-1}{\dfrac{a_{n+1}}{a_{n}}}-\cot^{-1}{\dfrac{a_{n}}{a_{n-1}}}$$ so $$\sum_{n=1}^{\infty}\cot^{-1}{a^2_{n}}=\cot^{-1}{(2+\sqrt{3})}=\dfrac{\pi}{12}$$ since $$\sum_{n=1}^{\infty}\cot^{-1}{a^2_{n}}=\cot^{-1}{a^2_{1}}+\lim_{N\to\infty}\sum_{n=2}^{N}\cot^{-1}{a^2_{n}}=\cot^{-1}{a^2_{1}}+\lim_{n\to\infty}\sum_{n=2}^{N}\left(\cot^{-1}{\dfrac{a_{n+1}}{a_{n}}}-\cot^{-1}{\dfrac{a_{n}}{a_{n-1}}}\right)$$ $$=\cot^{-1}{a^2_{1}}+\lim_{n\to\infty}\left(\cot^{-1}{\dfrac{a_{N+1}}{a_{N}}}-\cot^{-1}{\dfrac{a_{2}}{a_{1}}}\right)=\lim_{N\to\infty}\cot^{-1}{\dfrac{a_{N+1}}{a_{N}}}$$

let $a_{n+1}=4a_{n}-a_{n-1}$,we easy prove $$\lim_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}=r>1$$is exst. so $$r^2=4r-1\Longrightarrow r=2+\sqrt{3}$$