TITLE: Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator
QUESTION [8 upvotes]: 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!

REPLY [4 votes]: $$\tan^2\frac{\pi}{5}+\tan^2\frac{2\pi}{5}=\frac{1-\cos\frac{2\pi}{5}}{1+\cos\frac{2\pi}{5}}+\frac{1-\cos\frac{4\pi}{5}}{1+\cos\frac{4\pi}{5}}=$$
$$=\frac{2-2\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}{1+\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}=$$
$$=\frac{2-\frac{4\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}{2\sin\frac{2\pi}{5}}}{1+\frac{2\sin\frac{\pi}{5}\cos\frac{2\pi}{5}+2\sin\frac{\pi}{5}\cos\frac{4\pi}{5}}{2\sin\frac{\pi}{5}}+\frac{4\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}{4\sin\frac{2\pi}{5}}}=$$
$$=\frac{2-\frac{\sin\frac{8\pi}{5}}{2\sin\frac{2\pi}{5}}}{1+\frac{\sin\frac{3\pi}{5}-\sin\frac{\pi}{5}+\sin\frac{5\pi}{5}-\sin\frac{3\pi}{5}}{2\sin\frac{\pi}{5}}+\frac{\sin\frac{8\pi}{5}}{4\sin\frac{2\pi}{5}}}=\frac{2+\frac{1}{2}}{1-\frac{1}{2}-\frac{1}{4}}=10.$$