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 $\tan\left(\frac{\pi}{5}\right)=x$
$$\tan\left(\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}\right)=0$$
$$\implies S_1-S_3+S_5=0$$
($S_k$ represents sum of tangents taken $k$ at a time)
$$\implies 5x-10x^3+x^5=0$$
Now the roots of this equation are $\tan\left(\frac{\pi}{5}\right),\tan\left(\frac{2\pi}{5}\right),\tan\left(\frac{3\pi}{5}\right),\tan\left(\frac{4\pi}{5}\right),\tan\left(\frac{\pi}{5}\right)$. Now sum of squares of the roots is $$\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)+\tan^2\left(\frac{3\pi}{5}\right)+\tan^2\left(\frac{4\pi}{5}\right)+\tan^2\left(\frac{\pi}{5}\right)=2\left(\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)\right)$$ Now this when calculated from basic concept of theory of equations applying to given polynomial comes to be $20$.
Hence $$2\left(\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)\right)=20 \implies \tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)=10$$ 
