$\newcommand{\cs}{\operatorname{cs}}$ My actual question is at the bottom.
Let $$ a\diamond b = \frac{a+b}{1+ab} $$ and $$ a\dagger=\dfrac{1-a}{1+a}. $$ Then $$ a\dagger\dagger = a $$ and \begin{align} (ab)\dagger & = (a\dagger)\diamond (b\dagger), \\[8pt] (a\diamond b)\dagger & = (a\dagger)(b\dagger), \end{align} and hence $$ (a\diamond b)\diamond c = a\diamond (b\diamond c) $$ (and this comes to $\dfrac{a+b+c+abc}{1+ab+ac+bc}$).
If we then let $g(a) = a\diamond a$ and $f(a) = (a\dagger)\diamond (a\dagger)$, then we have \begin{align} g(a) & = f(a\dagger) \\[6pt] f(a) & = g(a\dagger), \end{align} and \begin{align} g\left(\frac 1 a\right) = g(a), & \qquad f\left(\frac 1 a\right) = -f(a), \\[6pt] g(-a) = -g(a), & \qquad f(-a) = f(a). \end{align} Then we have $$ \tan\frac\alpha2\cdot\tan\frac\beta2= \tan\frac\gamma2 \iff \cos\alpha\diamond\cos\beta = \cos\gamma. $$ Let us define the "stereographic cosine" of $a$ to be the $x$-coordinate of the projection of the point $(0,a)$ onto the circle $x^2+y^2=1$ along a line through the point $(-1,0)$. Then $$ \cs a = (a^2)\dagger = (a\dagger)\diamond(a\dagger) $$ and we have the identity $$ \cs(a_1\cdots a_n)=\cs a_1 \diamond\cdots\diamond\cs a_n. $$ (This identity will appear in a forthcoming publication.)
My question is whether this somewhat suprisingly complex set of identities, with some of the rhythm of trigonometric identities, following from this seemingly trivially simple set of definitions, has a literature? Has this seen the light of day somewhere?
