In another posting I wrote about a trigonometric relation I had derived, but that ended up not being the main point of the posting:


[Strange pattern in rounding errors?](https://mathoverflow.net/questions/116214/strange-pattern-in-rounding-errors)


So as long as we're here, let's make it the main point of *this* posting. I posted [something like this](https://math.stackexchange.com/questions/255390/referenence-request-yet-another-tangent-half-angle-formula) a couple of days ago on stackexchange, with no answers yet.


Let's make this *two* questions:


* Is this "known" in the sense of being in some book, refereed paper, or the like?;
* Is there a *straightforward* way to prove this?


Here's how I derived this relation: I showed that a certain function satisfies a certain differential equation; then I showed that a certain function emerges as the antiderivative that you get by the usual second-semester-calculus methods; then I said these *ought* to be the same thing because they both solve a geometry problem that arose in some amateur cartography of a (maybe?) somewhat impractical sort, therefore they must be the same; then I checked it numerically and it checked. **But there ought to be a more straightforward way.**


Here's the result: If $$\tan\gamma=\dfrac{\sin\alpha\sin\beta}{\cos\alpha+\cos\beta}$$ and $\alpha, \beta, \gamma\in(0,\pi)$ or $\alpha,\beta,\gamma\in(-\pi,0)$ then
$$
\tan\dfrac\gamma2=\tan\dfrac\alpha2\cdot\tan\dfrac\beta2.
$$


A tangent half-angle formula that everyone knows, or at least that's out there in trigonometry-for-adults books that were occasionally published before about 1930, says
$$
\frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta} = \tan\frac{\alpha+\beta}{2}.
$$
Does it make any sort of sense to say that the fact that what I derived, and this "known" identity are reminiscent of each other has some reason behind it? (So I guess this is really *three* questions.)



A "known" tangent half-angle formula?
Using the tangent double-angle formula $\tan\gamma=\frac{2\tan\tfrac{\gamma}{2}}{1-\tan^2\tfrac{\gamma}{2}}$ we get
$$\begin{align}
\tan\gamma & = \frac{2\tan\tfrac{\beta}{2}\tan\tfrac{\alpha}{2}}{1-\tan^2\tfrac{\beta}{2}\tan^2\tfrac{\alpha}{2}} \\[10pt]
& = \frac{2\sin\tfrac{\beta}{2}\sin\tfrac{\alpha}{2}\cos\tfrac{\beta}{2}\cos\tfrac{\alpha}{2}}{\cos^2\tfrac{\beta}{2}\cos^2\tfrac{\alpha}{2}-\sin^2\tfrac{\beta}{2}\sin^2\tfrac{\alpha}{2}} \\[10pt]
& =\frac{\sin\beta\sin\alpha}{2(\cos\tfrac{\beta}{2}\cos\tfrac{\alpha}{2}-\sin\tfrac{\beta}{2}\sin\tfrac{\alpha}{2})(\cos\tfrac{\beta}{2}\cos\tfrac{\alpha}{2}+\sin\tfrac{\beta}{2}\sin\tfrac{\alpha}{2})} \\[10pt]
& =\frac{\sin\beta\sin\alpha}{2\cos\tfrac{\beta+\alpha}{2}\cos\tfrac{\beta-\alpha}{2}} \\[10pt]
& =\frac{\sin\beta\sin\alpha}{\cos\beta+\cos\alpha}
\end{align}$$


