在△ABC中，已知sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=cos^2(B/2).求证cot(A/2).cot（B/2).cot（C/2)

sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=cos^2(B/2)
<=> ½* (cosA+cosC) - 2 sin^2(B/2)=0
<=> cos(A+C)/2 * cos(A-C)/2 - sin^2(B/2)= sin^2(B/2)
<=> sin(B/2)*[ cos(A-C)/2 - cos(A+C)/2] = sin^2(B/2)
<=> sin(B/2)*sin(A/2)*sin(C/2) = ½* sin^2(B/2)
<=> sin(B/2) = 2* sin(A/2)*sin(C/2)

cot(A/2) + cot(C/2)
=(cosA+1)/sinA + (cosC+1)/sinC
=[sin(A+C)+sinA+sinC]/(sinA*sinC)
=[sinA+sin B+sinC]/(sinA*sinC)
=4cos(A/2)*cos(B/2)*cos(C/2) /[4sin(A/2)*cos(A/2)*sin(C/2)*cos(C/2)]
=cos(B/2)/[sin(A/2)*sin(c/2)]
=2cos(B/2)/sin(B/2)
=2cot(B/2)
cot(A/2).cot（B/2).cot（C/2)成等差数列