Question

已知cos(a+b)=-1,tana=2,则tanb的值为___

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Answer 1

已知cos(α+β)=-1,tanα=2,则tanβ的值为___
解：∵cos(α+β)=-1，∴sin(α+β)=±√[1-cos²(α+β)]=0，tan(α+β)=sin(α+β)/cos(α+β)=0
故tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=(2+tanβ)/(1-2tanβ)=0
∴2+tanβ=0，即tanβ=-2.

Answer 2

tana=2
sina/cosa=2
sin^2a+cos^2a=1
cosa=1/根号5
sina=2/根号5

cos(a+b)=cosacoab-sinasianb=-1
1/根号5cosb-2/根号5sinb=-1
sin^2b+cos^2b=1
cosb=-1/根号5 sinb=2/根号5
tanb=sinb/cosb=-2

Answer 3

cos(a+b) = -1
sin(a+b)=0
tan(a+b)=0
tan b = tan [(a+b) - a]
=[ tan(a+b) - tana] / [ 1 + tan (a+b)*tana]
= -2