已知 sin(A+B)=1\2,sin(A-B)=1\3,求tan(A+B)-tanA-tanB\tanB*tanB*t
1个回答

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