∵2x²-（√3+1）x+m=0的两根为sinα,cosα
∴Δ=(√3+1)²-4m≥0
sinα+cosα=(√3+1)/2①,sinαcosα=m/2②
①²:1+2sinαcosα=(4+2√3)/4
∴ sinαcosα=√3/4=m/2
∴m=√3/2

∴sinα=√3/2,cosα=1/2,α=2kπ+π/3,k∈Z

或sinα=1/2,cosα=√3/2,α=2kπ+π/6 ,k∈Z

当α=2kπ+π/3,k∈Z
∴sinα/(1-cosα)+cosα/(1-tanα)
=(√3/2)/(1-1/2)+(1/2)/(1-√3)
=√3-(√3+1)/4=(3√3-1)/4

当α=2kπ+π/6 ,k∈Z
sinα/(1-cosα)+cosα/(1-tanα)
=（1/2)/(1-√3/2)+(√3/2)/(1-√3/3)
=2+√3+3/4*(√3+1)
=(7√3+11)/4