已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1天哪……谁会啊……
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![已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1天哪……谁会啊……](/uploads/image/z/576224-8-4.jpg?t=%E5%B7%B2%E7%9F%A5ABC%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E5%86%85%E8%A7%92%2C%E6%B1%82%E8%AF%81tanA%2F2%2AtanB%2F2%2BtanB%2F2%2AtanC%2F2%2BtanC%2F2%2AtanA%2F2%3D1%E5%A4%A9%E5%93%AA%E2%80%A6%E2%80%A6%E8%B0%81%E4%BC%9A%E5%95%8A%E2%80%A6%E2%80%A6)
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已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1天哪……谁会啊……
已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1
天哪……
谁会啊……
已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1天哪……谁会啊……
tanA/2×tanB/2+tanB/2×tanC/2+tanC/2×tanA/2
=tanA/2×tanB/2+tanC/2×(tanA/2+tanB/2)
=tanA/2×tanB/2+tan[90-(A+B)/2]×(tanA/2+tanB/2)
=tanA/2×tanB/2+cot(A/2+B/2)×(tanA/2+tanB/2)
=tanA/2×tanB/2+(tanA/2+tanB/2)/tan(A/2+B/2)
=tanA/2×tanB/2+1-tanA/2×tanB/2
=1