sinA/sin(A+B)=2/3,求tan(B/2)*cot(A+B/2)的值.

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∵sinA/sin(A+B)=2/3
∴3sinA=2sin(A+B)
∴3sin[(A+B/2)-B/2]=2sin[(A+B/2)+B/2]
3[sin(A+B/2)cosB/2-cos(A+B/2)sinB/2]=2[sin(A+B/2)cosB/2+cos(A+B/2)sinB/2]
sin(A+B/2)cosB/2=5cos(A+B/2)sinB/2
[cos(A+B/2)sinB/2]/[sin(A+B/2)cosB/2]=1/5
∴tan(B/2)*cot(A+B/2)=1/5

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