Question

已知tan(a+b)=1/3,tana=-2,求tanb

Answer 1

∵tan(a+b)=(tana+tanb)/(1-tanatanb)

∴tanb=[tan(a+b)-tana]/[tan(a+b)tana+1)

∵ tan(a+b)=1/3,tana=-2

∴tanb=[(1/3)-(-2)]/[(1/3)(-2)+1]=7/5