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si

问题补充：

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＝＿＿＿

答案：

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)/bc

cosA = (b^2+c^2-a^2)/2bc

1-cosA = (2bc - b^2 - c^2 +a^2)/2bc

sinA/(1-cosA) = 2/(1/2) = 4