已知A,B都是锐角,A+B不等于π/2,(1+tanA)(1+tanB)=2,求证A+B=π/4
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![已知A,B都是锐角,A+B不等于π/2,(1+tanA)(1+tanB)=2,求证A+B=π/4](/uploads/image/z/4024527-15-7.jpg?t=%E5%B7%B2%E7%9F%A5A%2CB%E9%83%BD%E6%98%AF%E9%94%90%E8%A7%92%2CA%2BB%E4%B8%8D%E7%AD%89%E4%BA%8E%CF%80%2F2%2C%281%2BtanA%29%281%2BtanB%29%3D2%2C%E6%B1%82%E8%AF%81A%2BB%3D%CF%80%2F4)
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已知A,B都是锐角,A+B不等于π/2,(1+tanA)(1+tanB)=2,求证A+B=π/4
已知A,B都是锐角,A+B不等于π/2,(1+tanA)(1+tanB)=2,求证A+B=π/4
已知A,B都是锐角,A+B不等于π/2,(1+tanA)(1+tanB)=2,求证A+B=π/4
(1+tanA)(1+tanB)=2
1+tanA+tanB+tanAtanB=2
tanA+tanB=1-tanAtanB
∴tan(A+B)=(tanA+tanB)/(1-tanA
tanB)=1=tanπ/4
∴A+B=π/4
(1+tanA)(1+tanB)=2,
乘开来:1+tanA+tanB+tanA*tanB=2
即:tanA+tanB+tanA*tanB=1
移项得 tanA+tanB=1-tanA*tanB
相除得 1=(tanA+tanB)/(1-tanA*tanB)
=tan(A+B)
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