$\newcommand{\cs}{\operatorname{cs}}$ 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?

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    $\begingroup$ Forgive me if this seems unnecessary. I'm interested in finding out where you get all these very interesting ideas from. What is your motivation? $\endgroup$ – user122283 May 16 '14 at 23:15
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    $\begingroup$ @SanathDevalapurkar : Thank you. I get them from fiddling with elementary geometry and trigonometry. $\endgroup$ – Michael Hardy May 17 '14 at 2:57
  • $\begingroup$ Sir, by forthcoming publication do you mean in a journal? Or preprint like Arxiv? $\endgroup$ – Kugelblitz Mar 11 '15 at 12:16
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    $\begingroup$ This material is somewhat related to the Gudermannian identities, which connect the circular functions with the hyperbolic functions without (explicit) use of complex numbers. $\endgroup$ – PM 2Ring Mar 11 '15 at 13:00

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