知tan(A-B)/tanA+sin^2C/sin^2A=1求证tanAtanB=tan^2C

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知tan(A-B)/tanA+sin^2C/sin^2A=1求证tanAtanB=tan^2C
知tan(A-B)/tanA+sin^2C/sin^2A=1求证tanAtanB=tan^2C

知tan(A-B)/tanA+sin^2C/sin^2A=1求证tanAtanB=tan^2C
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
化简,得
(tan²A*tanB+tanB)/tanA(1+tanA*tanB)=sin²C/sin²A
tanB*(sec²A)/tanA(1+tanA*tanB)=sin²C/sin²A
交叉相乘,得
tanB*tan²A=tanA(1+tanA*tanB)*sin²C
两边除以tanA
tanB*tanA=(1+tanA*tanB)*sin²C
左边做一个+1 -1动作,得
tanB*tanA+1-1=(1+tanA*tanB)*sin²C
1-1/(1+tanA*tanB) = sin²C
移项,得 1-sin²C=1/(1+tanA*tanB)
由于1-sin²C=cos²C cos²C=1/sec²C
得 1/sec²C=1/(1+tanA*tanB)
sec²C=1+tanA*tanB
sec²C-1=tanA*tanB
因为sec²C-1=tan²C
得 tanA*tanB=tan²C
证毕!

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