cos(3π/2+a)=3/5,a∈（π/2，π），tan(π+B)=1/2,求cot(a-2B)的值

∵cos(3π/2+a)=3/5
==>sina=3/5
又a∈（π/2，π）
∴cosa<0,即cosa=-√(1-sin²a)=-4/5
∴tana=-3/4
∵tan(π+B)=1/2==>tanB=1/2
∴tan(2B)=2tanB/(1-tan²B)
=2(1/2)/[1-(1/2)²]
=4/3
故 cot(a-2B)=1/tan(a-2B)
=[1+tana*tan(2B)]/[tana-tan(2B)]
=[1+(-3/4)(4/3)]/[(-3/4)-(4/3)]
=0.