I stumbled on a nice identity earlier on: $\tan(50^\circ)\tan(60^\circ)\tan(70^\circ) = \tan(80^\circ)$. It’s one of those I can prove, but don’t yet understand.

My proof goes along the lines of, let $T = \tan(10^\circ)$. Then the LHS is $\tan((60-10)^\circ)\tan(60^\circ)\tan((60+10)^\circ)$, which works out to $\frac{\sqrt{3}-T}{1+T\sqrt{3}} \cdot \sqrt{3}\cdot \frac{\sqrt{3}+T}{1-T\sqrt{3}}$.

Simplifying that gives a left-hand side of $\sqrt{3} \cdot \frac{3-T^2}{1-3T^2}$

Now the clever bit: if $\tan(\theta) = t$, then $\tan(3\theta) = \frac{3t - t^3}{1-3t^2}$, In particular, that means $\frac{3T - T^3}{1-3T^2} = \tan(30^\circ) = \frac{1}{\sqrt{3}}$.

That’s very close to the LHS. If we write the LHS as $\frac{\sqrt{3}}{T}\cdot \frac{3T - T^3}{1-3T^2}$, we can replace the second factor with $\frac{1}{\sqrt{3}}$ and hey! It all drops out to $\frac{1}{T}$.

And $\frac{1}{\tan(10^\circ)} = \tan(80^\circ)$ $\blacksquare$

That’s fine, it’s a neat and solid proof, but it’s unsatisfactory to me, and I’ve not been able to draw a Convincing Picture. If you’ve got any ideas, I’d love to hear them!