已知角A、角B和角C是三角形ABC的内角,求证求证:tan2/A*tan2/B+tan2/B*tan2/C+tan2/C*tan2/A=1\
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![已知角A、角B和角C是三角形ABC的内角,求证求证:tan2/A*tan2/B+tan2/B*tan2/C+tan2/C*tan2/A=1\](/uploads/image/z/5326505-17-5.jpg?t=%E5%B7%B2%E7%9F%A5%E8%A7%92A%E3%80%81%E8%A7%92B%E5%92%8C%E8%A7%92C%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%86%85%E8%A7%92%2C%E6%B1%82%E8%AF%81%E6%B1%82%E8%AF%81%EF%BC%9Atan2%2FA%2Atan2%2FB%2Btan2%2FB%2Atan2%2FC%2Btan2%2FC%2Atan2%2FA%3D1%5C)
已知角A、角B和角C是三角形ABC的内角,求证求证:tan2/A*tan2/B+tan2/B*tan2/C+tan2/C*tan2/A=1\
已知角A、角B和角C是三角形ABC的内角,求证
求证:tan2/A*tan2/B+tan2/B*tan2/C+tan2/C*tan2/A=1\
已知角A、角B和角C是三角形ABC的内角,求证求证:tan2/A*tan2/B+tan2/B*tan2/C+tan2/C*tan2/A=1\
原式=tan(A/2)*[tan(B/2)+tan(C/2)]+tan(B/2)*tan(C/2)
=tan(A/2)*[tan(B+C)/2]*[1-tan(B/2)*tan(C/2)]+tan(B/2)*tan(C/2)
其中:应用公式:tanα+tanβ=tan(α+β)*(1-tanα*tanβ)
=tan(A/2)*cot(A/2)[1-tan(B/2)*tan(C/2)]+tan(B/2)*tan(C/2)
=1
A+B+C=180
C=180-A-B
代入tan2/A*tan2/B+tan2/B*tan2/C+tan2/C*tan2/A此式
然后整理一下,就可以得到了