Spherical Trig question

Refer to the following image: https://upload.wikimedia.org/wikipedia/commons/thumb/f/fa/Pentagramma-mirificum.png/275px-Pentagramma-mirificum.png


In Gauss's notation, one has $\tan^2(a) = \epsilon^{-1}, \tan^2(b) = \beta^{-1}, \tan^2 (c) = \alpha, \tan^2(A) = \delta, \tan^2(B) = \gamma$. Note that by Gauss's formulas, one has $1 + \beta = \delta \epsilon$, implying that $\beta = \delta \epsilon - 1$. Furthermore, one has $1 + \epsilon = \beta \gamma$, implying that $\epsilon = \beta\gamma - 1$. Together this implies that $\beta = \delta \beta \gamma - \delta \ - 1$, or equivalently, $\beta = \frac{\delta + 1}{\delta\gamma - 1}$. Substituting back yields

$$b = \tan^{-1}\left(\sqrt{\frac{\tan^2(A) \tan^2(B) - 1}{\tan^2(A) + 1}}\right)$$