已知:tan(A-B)\tanA+(sinC/sinA)*(sinC/sinA)=1,求证tanA*tanB=tanC*tanC

问题描述:

已知:tan(A-B)\tanA+(sinC/sinA)*(sinC/sinA)=1,求证tanA*tanB=tanC*tanC

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)tan(A-B)/tanA+sin²C/sin²A=1 左右移项得 1-[(tanA-tanB)/(1+tanA*tanB)]/tanA=sin²C/sin²A 左边化简一下得 (tan²A*tanB+tanB)/tanA(1+tanA*tanB)=sin&...