tan(a-b/2)=0.5 tan(b-a/2)=-1/3求tan(a+b)

∵ tan(α+β)=(tanα+tanβ)/(1-tanα·tanβ)

∴
tan(a+b)=tan2(a-b/2+b-a/2)
=(tan(a-b)+tan(a+b))/(1-tan(a-b)tan(a+b))

tan(a+b)/2=
=(tan(a-b/2)+tan(b-a/2))/(1-tan(a-b/2)tan(b-a/2))
=(0.5-1/3)/(1+0.5*1/3)
=1/7

∵an2α＝2tanα/（1－tan^2α ）

∴
tan(a+b)
=(2/7)/(1-1/49)
=7/24
∴
tan(a+b)=7/24

tan[(a-b/2)+(b-a/2)]=tan[(a+b)/2]
设(a-b/2)=A，(b-a/2)=B
tan[(a+b)/2]=tan（A+B）=(tanA+tanB)/(1-tanA*tanB)
=1/7
tan(a+b)=tan[2*(a+b/2)],设(a+b/2)=C，tanC=1/7
=2tanC/(1+tanC^2)
=7/25

tan((a-b/2)+(b-a/2))=(tan(a-b/2)+tan(b-a/2))/(1-tan(a-b/2)tan(b-a/2))
即tan(a/2+b/2)=(0.5-1/3)/(1+0.5*(1/3))=1/7
tan(a+b)=2*tan(a/2+b/2)/(1-tan(a/2+b/2)^2)=(2/7)/(1-1/49)=7/24

tan[（a＋b）/2]=tan[（a－b/2）＋（b－a/2）]=[tan（a－b/2）＋tan（b－a/2）]/[1－tan(a－b/2)tan(b－a/2)]＝1/7
tan(a＋b)=2tan[（a＋b）/2]/[1－{tan[(a＋b)/2]}∧2=7/24。

tan[(a+b)/2]=tan(a-b/2 + b-a/2)= [tan(a-b/2) + tan(b-a/2)]/[1-t