donc, ajoutant ensemble, on aura
![{\displaystyle \mathrm {\left(\operatorname {tang} {\frac {A}{2}}+\operatorname {tang} {\frac {B}{2}}\right)\operatorname {tang} {\frac {C}{2}}} ={\frac {\sin {\cfrac {a-b+c}{2}}+\sin {\cfrac {b+c-a}{2}}}{\sin {\cfrac {a+b+c}{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0577f365653ad3518ede518f724651589d072d51)
![{\displaystyle ={\frac {2\sin {\cfrac {c}{2}}\cos {\cfrac {a-b}{2}}}{\sin {\cfrac {a+b+c}{2}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eddc65f336938063c8573d20cd43e9b59334baf5)
et, retranchant les mêmes équations l’une de l’autre, on aura
![{\displaystyle \mathrm {\left(\operatorname {tang} {\frac {A}{2}}-\operatorname {tang} {\frac {B}{2}}\right)\operatorname {tang} {\frac {C}{2}}} ={\frac {\sin {\cfrac {a-b+c}{2}}-\sin {\cfrac {b+c-a}{2}}}{\sin {\cfrac {a+b+c}{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1fcdca0616be2a9f63ea53453c4bec20c612333)
![{\displaystyle ={\frac {2\cos {\cfrac {c}{2}}\sin {\cfrac {a-b}{2}}}{\sin {\cfrac {a+b+c}{2}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c3a8c069847f420ca210f58cf8e40ddadbc79b3)
D’un autre côté, on a
![{\displaystyle 1+\mathrm {\operatorname {tang} {\frac {A}{2}}\operatorname {tang} {\frac {B}{2}}} ={\frac {\sin {\cfrac {a+b+c}{2}}+\sin {\cfrac {a+b-c}{2}}}{\sin {\cfrac {a+b+c}{2}}}}={\frac {2\sin {\cfrac {a+b}{2}}\cos {\cfrac {c}{2}}}{\sin {\cfrac {a+b+c}{2}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f1953a940c9ec4ea1ffd4b17f8678994c034aa5)
![{\displaystyle 1-\mathrm {\operatorname {tang} {\frac {A}{2}}\operatorname {tang} {\frac {B}{2}}} ={\frac {\sin {\cfrac {a+b+c}{2}}-\sin {\cfrac {a+b-c}{2}}}{\sin {\cfrac {a+b+c}{2}}}}={\frac {2\sin {\cfrac {c}{2}}\cos {\cfrac {a+b}{2}}}{\sin {\cfrac {a+b+c}{2}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3273079ca8d5604454885b612af9a74933b22ac6)
Donc, puisque
on aura ces deux équations
![{\displaystyle (r)\qquad \qquad \qquad \qquad \mathrm {\operatorname {tang} {\frac {A+B}{2}}\operatorname {tang} {\frac {C}{2}}} ={\frac {\cos {\cfrac {a-b}{2}}}{\cos {\cfrac {a+b}{2}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8d0c315e431a0d3558465815c95c1dbed872c1a)
![{\displaystyle (s)\qquad \qquad \qquad \qquad \mathrm {\operatorname {tang} {\frac {A-B}{2}}\operatorname {tang} {\frac {C}{2}}} ={\frac {\sin {\cfrac {a-b}{2}}}{\sin {\cfrac {a+b}{2}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/787745438f80e6fcbd0024bce366df2c26c08576)
On peut déduire des formules semblables de l’équation
du no 24, et, sans faire un nouveau calcul, il n’y aura qu’à changer les côtés
dans les suppléments à deux droits des angles
et ces angles