设tanα=1/2，tanβ=1/3，则cot（α+2β）的值

tan(A+B) = (tanA+tanB)/(1-tanAtanB)

所以tan2β=3/4

tan（α+2β）=2，

cot（α+2β）=1/tan（α+2β）=1/2

assume alpha=a, belta=b
tana=1/2 tanb=1/3 tan2b=(2tanb)/(1-(tanb)^2 (^2 means square)
tan(a+2b)=(tana+tan2b)/(1-tana*tan2b)
tan(a+2b)=1/cot(a+2b)
then you can get the right answer guys.