Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $\tan{\left(\frac{2\pi}{5}\right)}$ to find that the answer is $10$. However, I want to know whether there is a faster way that does not involve calculating those values.
So far, I have this: let $a=\tan{\left(\frac{\pi}{5}\right)}$ and $b=\tan{\left(\frac{2\pi}{5}\right)}$; then, $b=\frac{2a}{1-a^2}$ and $a=-\frac{2b}{1-b^2}$, so multiplying the two and simplifying gives:
$$a^2+b^2={\left(ab\right)}^2+5$$
Any ideas? Thanks!

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 $\color{red}{10}$.
