题目

已知sin(a+b)=2/3,sin(a-b)=3/4,求[tan(a+b）-tana-tanb]/tan²b*tan(a+b)的值

提问时间：2020-11-22

答案

sin(a+b)=sinacosb+cosasinb=2/3……(1)
sin(a-b)=sinacosb-cosasinb=3/4……(2)
(1)+(2)
2sinacosb=17/12,
sinacosb=17/24
(1)-(2)
2cosasinb=-1/12,
cosasinb=-1/24
tan(a+b)-tana-tanb
=tan(a+b)-(tana+tanb)
=tan(a+b)-[tan(a+b)*(1-tanatanb]
=tana*tanb*tan(a+b)
(tan(a+b)-tana-tanb)/(tan^2b*tan(a+b))
=tana*tanb*tan(a+b)/(tan^2b*tan(a+b))
=tana*tanb/tan^2b
=tana/tanb
=sinacosb/cosasinb
=(17/24)/(-1/24)
=-17