设a,b,c分别是三角形ABC的三个内角A,B,C所对的边,S△ABC=a^-(b-C)^2,则sinA/1-cosA＝＿＿＿

S△ABC = 1/2 bc sinA所以 1/2 bc sinA = (a^2 -(b-C)^2)sinA = 2(a^2 -b^2 -c^2 +2bc)/bccosA = (b^2+c^2-a^2)/2bc1-cosA = (2bc - b^2 - c^2 +a^2)/2bcsinA/(1-cosA) = 2/(1/2) = 4