题目

已知 sin(A+B)=12,sin(A-B)=13,求tan(A+B)-tanA-tanB anB*tanB*tan(A+B).

提问时间：2020-10-15

答案

sin(a+b)=sinacosb+cosasinb=1/2……(1)
sin(a-b)=sinacosb-cosasinb=1/3……(2)
(1)+(2)
2sinacosb=5/6,
sinacosb=5/12
(1)-(2)
2cosasinb=1/6,
cosasinb=1/12
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
=(5/12)/(1/12)
=5