已知A+B=5派/4,A ,B≠k派+派/2(k ∈Z ),求证(1+tan A )(1+tanB=2
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![已知A+B=5派/4,A ,B≠k派+派/2(k ∈Z ),求证(1+tan A )(1+tanB=2](/uploads/image/z/1649403-27-3.jpg?t=%E5%B7%B2%E7%9F%A5A%2BB%EF%BC%9D5%E6%B4%BE%2F4%2CA+%2CB%E2%89%A0k%E6%B4%BE%2B%E6%B4%BE%2F2%EF%BC%88k+%E2%88%88Z+%EF%BC%89%2C%E6%B1%82%E8%AF%81%EF%BC%881%2Btan+A+%EF%BC%89%EF%BC%881%2BtanB%EF%BC%9D2)
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已知A+B=5派/4,A ,B≠k派+派/2(k ∈Z ),求证(1+tan A )(1+tanB=2
已知A+B=5派/4,A ,B≠k派+派/2(k ∈Z ),求证(1+tan A )(1+tanB=2
已知A+B=5派/4,A ,B≠k派+派/2(k ∈Z ),求证(1+tan A )(1+tanB=2
证明:
tan( A+B)=(tan A +tanB)/(1-tan A*tanB)=tan 5π/4=1【第三象限且A ,B≠kπ+π/2(k ∈Z )】
则tan A +tanB=1-tan A*tanB
移项得tan A +tanB+tan A*tanB=1
而(1+tan A )(1+tanB)=1+tan A +tanB+tan A*tanB=1+1=2
得证