"Recall" that $$ \frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta} = \cos \gamma \quad \text{ if OR only if (I forgot which) } \quad \tan\frac\alpha2\cdot\tan\frac\beta 2 = \tan\frac\gamma 2 $$ (where the "or" is inclusive). (I never knew this identity before I derived it myself, so you can construe the word "recall" in the sense of Plato.) PS: There should be some absolute value signs there.
In this answer by Lukas Geyer, we see that $$ \int_0^u \frac{dt}{R+\cos t} = \frac 2 {\sqrt{R^2-1}} \arctan\left( \tan \frac u 2 \cdot \sqrt{ \frac{R-1}{R+1}} \ \right). $$ (He stops short of saying that explicitly except in the case where $u=2\pi$.)
Two times the arctangent of a product of tangents is what $\gamma$ is in the trigonometric identity above. So we ask: what if the radical $\sqrt{\dfrac{R-1}{R+1}}$ were $\tan\dfrac\beta2$ for some $\beta$? It that were true, then we would have $$ \sqrt{\frac{R-1}{R+1}} = \tan\frac \beta 2; \quad\text{therefore} \quad \tan\beta = \sqrt{R^2-1}, $$ Letting $I$ be the integral above, we get $$ \tan\left( \frac {I \tan\beta} 2 \right) = \tan\frac u 2\cdot \tan\frac\beta 2. $$ This looks so weird that I am suspicious. Does this look anything like any standard result that I've never heard of? Is something of interest known about this?
