sinacosa/(1-cos2a)=1,tan(a-B)=-2/3,则tan(B-2a)=?

sinacosa/(1-cos2a)=1,sinacosa=1-cos2a=1-(1-2sin^2a)
sinacosa=2sin^2a
sina=0或 cosa=2sina
(1) sina=0 a=kπ
tan(a-B)=-tanB=-2/3
tanB=2/3
tan(B-2a)=tan(-2kπ+B)=tanB=2/3
(2) cosa=2sina
tana=1/2 tan2a=2tana/(1-tan^2a)=4/3
tan(a-B)
=(tana-tanB)/(1+tanatanB)
=(1/2-tanB)/(1+tanB/2)
=-2/3
tanB=7/4
tan(B-2a)
=(tanB-tan2a)/(1+tanBtan2a)
=(7/4-4/3)/(1+7/4*3/4)
=1/(1+21/16)
=16/37sorry 答案是1/8tan(B-2a)=(tanB-tan2a)/(1+tanBtan2a)=(7/4-4/3)/(1+7/4*4/3)=(21-16)/(12+28)=5/40=1/8嘎嘎 那2/3的答案是怎么排除的呢是有误1-cos2a≠01-cos2a=2sin^2a≠0所以sina≠0所以（1）应该舍去，只有1/8 sorry