(-2cos2C/(1+tanC))+1=-2(cosCcosC-sinCsinC)/(1+sinC/cosC)+1
=-2cosC(cosC-sinC)+1=sin2C-cos2C=√2sin(2C-π/4)
C属于(0,π/2) 故2C-π/4属于(-π/4,3π/4)
故当2C-π/4=π/2时,(-2cos2C/(1+tanC))+1有最大值√2
当2C-π/4=-π/4时,(-2cos2C/(1+tanC))+1有最小值-1
综合得到(-2cos2C/(1+tanC))+1的取值范围是(-1,√2]