14
PROBLÈMES.
de ces deux fractions la forme connue de binôme, savoir
Dans cette vue, nous ferons, pour abréger,
![{\displaystyle {\begin{aligned}&a=\operatorname {Cos} .A\operatorname {Cot} .B,&&a'=\operatorname {Sin} ,A\operatorname {Cot} .B,\\&b=\operatorname {Cos} .A'\operatorname {Cot} .B',&&b'=\operatorname {Sin} .A'\operatorname {Cot} .B',\\&c=\operatorname {Cos} .A''\operatorname {Cot} .B'',&&c'=\operatorname {Sin} .A''\operatorname {Cot} .B'',\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7954e839749b52b63280fd9f3148552a80b781e4)
![{\displaystyle {\begin{aligned}m=&\operatorname {Sin} .(\theta -A)\operatorname {Cot} .B\operatorname {Sin} .\theta '-\operatorname {Sin} .(\theta '-A')\operatorname {Cot} .B'\operatorname {Sin} .\theta ,\\n=&\operatorname {Sin} .(\theta -A)\operatorname {Cot} .B\operatorname {Sin} .\theta ''-\operatorname {Sin} .(\theta ''-A'')\operatorname {Cot} .B''\operatorname {Sin} .\theta ',\\o=&\operatorname {Sin} .(\theta '-A')\operatorname {Cot} .B'\operatorname {Sin} .\theta ''-\operatorname {Sin} .(\theta ''-A'')\operatorname {Cot} .B''\operatorname {Sin} .\theta \,;\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6193e49f1d4e2a3b7f472295da2e581aede40f7)
![{\displaystyle {\begin{aligned}m'=&\operatorname {Sin} .(\theta -A)\operatorname {Cot} .B\operatorname {Cos} .\theta '-\operatorname {Sin} .(\theta '-A')\operatorname {Cot} .B'\operatorname {Cos} .\theta ,\\n'=&\operatorname {Sin} .(\theta -A)\operatorname {Cot} .B\operatorname {Cos} .\theta ''-\operatorname {Sin} .(\theta ''-A'')\operatorname {Cot} .B''\operatorname {Cos} .\theta ,\\o'=&\operatorname {Sin} .(\theta '-A')\operatorname {Cot} .B'\operatorname {Cos} .\theta ''-\operatorname {Sin} .(\theta ''-A'')\operatorname {Cot} .B''\operatorname {Cos} .\theta '.\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c7f8f3a7dac3bfcdcc7b565056f2d7229ee4b34)
91. Enfin, proposons la dernière notation que la nature du problème exige, et qui paraît nécessaire pour présenter l’inconnue sous la forme la plus simple ; savoir,
![{\displaystyle {\begin{aligned}D=&hm-tn-c'h\operatorname {Sin} .t+b't\operatorname {Sin} .h,\\E=&hm'-tn'-ch\operatorname {Sin} .t+bt\operatorname {Sin} .h,\\F=&b'tn-c'hm,\\G=&b'tn'+btn-c'hm'-chm,\\H=&btn'-chm'\,;\\D'=&ho-t'n-a'h\operatorname {Sin} .t'+b't'\operatorname {Sin} .h,\\E'=&ho'-t'n'-ah\operatorname {Sin} .t'+bt'\operatorname {Sin} .h,\\F'=&b't'n-a'ho,\\G'=&bt'n+b't'n'-aho-a'ho',\\H'=&bt'n'-aho'.\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/492972b68f0906615c1f95e7b391a7a0c9cfc202)
92. Les deux expressions (89) deviendront alors