已知tan(a+b)=7,tan a *tan b =2/3 则cos(a-b)的值等于什么?

tana+tanb
=tan(a+b)*(1-tanatanb)
=7*(1-2/3)
=7/3
(tana-tanb)^2
=(tana+tanb)^2-4tanatanb
=49/9-8/3
=25/9
tan(a-b)=(tana-tanb)/(1+tanatanb)
[tan(a-b)]^2
=(tana-tanb)^2/(1+tanatanb)^2
=(25/9)^2/(1+2/3)^2
=25/9
1+[tan(a-b)]^2=1/[cos(a-b)]^2
1+25/9=1/[cos(a-b)]^2
cos(a-b)=±3/√34
我的才是对的

解:由tan(a+b)=7,tana*tanb=2/3~(1)
tan(a+b)=(tana+tanb)/(1-tana*tanb)
得tana+tanb=7/3~(2),
联立(1)(2)解得:
tana=1/3,tanb=2,或tana=2,tanb=1/3,
所以tan(A-B)=(tana-tanb)/(1+tana*tanb)=1或1.
所以cos(A-B)=正负根号2/2

tan(a+b)=(tana+tanb)/(1-tan a *tan b)=(tana+tanb)/(1-2/3)=7
则(tana+tanb)=21,
cos(a-b)=cosacosb+sinasinb,
sin(a+b)=sinacosb+cosasinb,
做比得(1+tanatanb)/(tana+tanb)=5/24
cos(a-b)=5/24*sin(a+b)=5/24*7/√50=7√2/48

tana+tanb
=tan(a+b)*(1-tanatanb)
=7*(1-2/3)
=7/3
(tana-tanb)^2
=(tana+tanb)^2-4tanatanb
=49/9-8/3
=25/9
tan(a-b)=(tana-tanb)/(1+tanatanb)
[tan(a-b)]^2
=(tana-tanb)^2/(1+tanatanb)^2
=(25/9)^2/(1+2/3)^2
=25/9
1+[tan(a-b)]^2=1/[cos(a-b)]^2
1+25/9=1/[cos(a-b)]^2
cos(a-b)=±3/√34

tan(a+b)=(tana+tanb)/(1-tana*tanb)=7
代入tan a *tan b =2/3,得到tana+tanb=7/3
可以解得tana=2,tanb=1/3 (或tana=1/3,tanb=12）
则tan(a-b)=(tana-tanb)/(1+tana*tanb)=正负1
则cos(a-b)=正负2分之根号2