Mathematics Diary posted the following identity this morning. If a + b + c = π then

tan(a) + tan(b) + tan(c) = tan(a) tan(b) tan(c).

I’d never seen that before. It’s striking that the sum of three numbers would also equal their product. In fact, the only way for the product of three numbers to be equal to their sum is for the three numbers to be tangents of angles that add up to π radians. I’ll explain below why that’s true.

First, we can generalize the identity above slightly by saying it holds if a + b + c is a multiple of π. In fact, it can be shown that

tan(a) + tan(b) + tan(c) = tan(a) tan(b) tan(c)

if and only if a + b + c is a multiple of π. (Here’s a sketch of a proof.)

Now suppose x + y + z = x y z. We can find numbers a, b, and c such that x = tan(a), y = tan(b), and z = tan(c) and it follows that a + b + c must be a multiple of π. But can we chose a, b, and c so that their sum is not just a multiple of π but exactly π? Yes. If a + b + c equaled kπ for some integer k, pick a new value of c equal to the original c minus (k-1)π. This leaves the value of tan(c) unchanged since the tangent function has period π.