tan（a+β）=5,tana=2,求tanβ

tan（a+β）=5,tana=2,求tanβ
tan（a+β）=5,tana=2,求tanβ

tan（a+β）=5,tana=2,求tanβ
tan(a+b)=5
(tana+tanb)/(1-tanatanb)=5
(2+tanb)/(1-2tanb)=5
2+tanb=5-10tanb
11tanb=3
tanb=3/11

tan（a+β）=5
(tanα+tanβ)/(1-tanαtanβ) = 5
tana=2
(2+tanβ)/(1-2tanβ) = 5
2+tanβ = 5-10tanβ
11tanβ = 3
tanβ = 3/11

tanβ=tan[(a+β)-a]=[tan(a+β)-tana]/[1-tan(a+β)tana]=(5-2)/(1-5×2)=1/3

tanβ=tan[(a+β)-a]
=[tan(a+β)-tana]/[1+tan(a+β)tana]
=(5-2)/(1+5×2)=3/11

tan（a+β）=(tana+tanβ)/(1-tanatanβ)=5, tana+tanβ=5-5tanatanβ
tanβ=(5-tana)/(1+5tana)=3/11