已知sin(α/2)-cos(α/2)=(√10)/5,α∈(π/2,π),tan(π-2β)的值

应该是求tan(π-2α)吧
tan(π-2α)= -tan(2α)= -2tanα/[1-(tanα)^2] 只需求tanα
已知式平方得:
1-sinα=2/5 sinα= 0.6
α∈(π/2,π) 所以cosα结果是24/7

,tan(π-b)=1/2,求tan(a-2b)的值?
sina/2-cosa/2=√10/5
sina/2*cosπ/4-cosa/2*sinπ/4=√5/5
sin(a/2-π/4)=√5/5
因为a属于(π/2,π),所以a/2-π/4属于(0,π/4)
所以cos(a/2-π/4)>0
cos(a/2-π/4)=√(1-sin²(a/2-π/4))=2√5/5
tan(a/2-π/4)=sin(a/2-π/4)/cos(a/2-π/4)=1/2
tan(a-π/2)=2*1/2/(1-(1/2)²)=4/3
tan(a-π/2)=-tan(π/2-a)=-cota=4/3
cota=-4/3
tana=-3/4
tan(π-b)=1/2
-tanb=1/2
tanb=-1/2
tab2b=2*(-1/2)/(1-(-1/2)²)=-4/3
tan(a-2b)=(-3/4-(-4/3))(1+(-3/4)(-4/3))
=7/24