（1+cotA)(1+cotC)
=[(cosA+sinA)*(cosC+sinC)]/(cosAcosC)
=(cosAcosC+cosAsinC+sinAcosC+sinAsinC)/(cosAcosC)
=[cosAcosC+sinAsinC+sin(A+C)]/(cosAcosC)=2
故cosAcosC+sinAsinC+sin(A+C)=2cosAcosC
即sin(A+C)=cosAcosC-sinAsinC=cos(A+C)
即sinB=-cosB
故sinB+cosB=0
故√2sin(B+π/4)=0
又B+π/4∈(π/4,5π/4)
故B+π/4=π
即B=3π/4
故sinB=√2/2
故log2sinB=log2(√2/2)]=-1/2