在△ABC中角A的正切等于(角B的正切+角C的正切)/(角B的余切+角C的余切),判断三角形

sinA＝(sinB+sinC)/(cosB+cosC)sinA-(sinB+sinC)/(cosB+cosC)＝0 sinA-2sin[(B+C)/2]cos[(B-C)/2]/2cos[(B+C)/2]cos[(B-C)/2]＝0 sinA- sin[(B+C)/2]/cos[(B+C)/2]＝0 2sin(A/2)cos(A/2)-cos(A/2)/sin(A/2)＝0 [2sin...