This question already has an answer here:

Strangest thing...*:

$$\tan^2 10^\circ+\tan^2 50^\circ+\tan^2 70^\circ=9\tag{1}$$

The trick, as always, is how to prove it.

My idea was to add a "missing" tangent and analyze a similar expression:

$$\tan^2 10^\circ+\tan^2 30^\circ+\tan^2 50^\circ+\tan^2 70^\circ$$

...and then to attack this sum pairwise (first and the last term, second and third). Despite the fact that I got the same angle ($80^\circ$) here and there, I got pretty much nowhere with this approach.

The other interesting fact is that (1) can be rewritten as:

$$\cot^2 20^\circ+\cot^2 40^\circ+\cot^2 80^\circ\tag{1}$$

...and now the angles are in nice geometric progression. That's the vector of attack that I'm trying to exploit now, but maybe you can entertain youself a little bit too.

*Borrowed from "Usual suspects"


marked as duplicate by mathlove, Toby Mak, N. F. Taussig, Jens, Community Nov 24 '18 at 14:18

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  • 3
    $\begingroup$ This is a very similar question. $\endgroup$ – Toby Mak Nov 24 '18 at 12:43

You can write it as $$\cot^220^\circ+\cot^240^\circ+\cot^260^\circ+\cdots+\cot^2160^\circ=\frac{56}3$$ That has eight multiples of $180^\circ/9$, and you can find a similar equation for other numbers instead of $9$.


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