已知角A角B角C是三角形ABC的内角,求证,tan(A/2)*tan(B/2)+tan(B/2)*tan(C/2)+ta_数学解答

已知角A角B角C是三角形ABC的内角,求证,tan(A/2)*tan(B/2)+tan(B/2)*tan(C/2)+ta

人气：366 ℃　时间：2020-06-23 00:24:54

解答

tan(90-A/2)=tan(B/2+C/2)=[tan(B/2) + tan(C/2)]/[1 - tan(B/2)tan(C/2)]所以1 - tan(B/2)tan(C/2) = [tan(B/2) + tan(C/2)]/tan(90-A/2)= [tan(B/2) + tan(C/2)]tan(A/2)所以tan(A/2)tan(B/2) + tan(B/2)tan(C/2) ...