已知sin(A+B)=2/3,sin(A-B)=3/4,求(tan(a+b)-tana-tanb)/(tan^2b*tan(a+b))

题目

已知sin(A+B)=2/3,sin(A-B)=3/4,求(tan(a+b)-tana-tanb)/(tan^2b*tan(a+b))
帮下忙

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 (3)
(1)-(2)
2cosAsinB=-1/12,cosAsinB=-1/24 (4)
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
=(3)/(4)
=-17