Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator

Let $a=\tan(\frac{\pi}{5})$ and $b= \tan(\frac{2 \pi}{5})$. Also let $s=a^2+b^2$ and $p=(a b)^2$. We have \begin{eqnarray*} a^2+b^2 &=&(ab)^2+5 \\ s&=&p+5. \end{eqnarray*} Now $a(1-b^2)=-2b$ and $b(1-a^2)=2a$ square these equations and add them together \begin{eqnarray*} 3(a^2+b^2)+4(ab)^2&=&(ab)^2(a^2+b^2) \\ 3s+4p&=&sp. \end{eqnarray*} Now eliminate $p$ and we have the quadratic $s^2-12s-20=0$. This has roots $2$ and $\color{red}{10}$.

Given this article here, your problem would come out to $2\cdot4+2=10$. You should check the article out because it gives some links to other pages and possibly proofs.

My question here has a proof of the identity.

Inspired/ Trying to make sense of Jaideep's Answer ... \begin{eqnarray*} \tan(5 \theta) =\frac{5 \tan(\theta)-10 \tan^3(\theta)+\tan^5(\theta)}{1-10\tan^2(\theta) +5\tan^4(\theta)} \end{eqnarray*} Let $\theta=\frac{\pi}{5}$ and $x=tan(\frac{\pi}{5})$. We have the quintic \begin{eqnarray*} 5x-10x^3+x^5=0 \end{eqnarray*} This has roots $tan(\frac{\pi}{5}),tan(\frac{2\pi}{5}),tan(\frac{3\pi}{5}),tan(\frac{4\pi}{5}),tan(\frac{5\pi}{5})$.

Note that $tan(\frac{3\pi}{5})=-tan(\frac{2\pi}{5})$ , $tan(\frac{4\pi}{5})=-tan(\frac{\pi}{5}) $ and $tan(\frac{5\pi}{5})=0$.

Now square this quintic & substitute $y=x^2$ and we have \begin{eqnarray*} 25y-100y^2+110y^3-\color{red}{20}y^4+y^5=0 \end{eqnarray*} The sum of the roots of this polynomial gives $2 (tan^2(\frac{\pi}{5})+tan^2(\frac{2\pi}{5}))=20$.