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

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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,a∈(π/2,π),tan(π+B)=1/2,求cot(a-2B)的值

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)
∴cosatanB=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.