三角形abc中,求证:tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1RTRTRTRTRTRT
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![三角形abc中,求证:tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1RTRTRTRTRTRT](/uploads/image/z/5191699-67-9.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2abc%E4%B8%AD%2C%E6%B1%82%E8%AF%81%3Atan%28A%2F2%29tan%28B%2F2%29%2Btan%28B%2F2%29tan%28C%2F2%29%2Btan%28C%2F2%29tan%28A%2F2%29%3D1RTRTRTRTRTRT)
三角形abc中,求证:tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1RTRTRTRTRTRT
三角形abc中,求证:tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1
RTRTRTRTRTRT
三角形abc中,求证:tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1RTRTRTRTRTRT
原式=tan(A/2)*
=tan(A/2)*tan<(B+C)/2>*<1-tan(B/2)*tan(C/2)>+tan(B/2)*tan(C/2)
=tan(A/2)*cot(A/2)*<1-tan(B/2)*tan(C/2)>+tan(B/2)*tan(C/2)
=1
应用公式:tanA+tanB=tan(A+B)*(1-tanA*tanB)
中括号我不会打,用的是<>
A/2=(180-B-C)/2=90-(B+C)/2
所以tan(A/2)=tan[90-(B+C)/2]
=cot(B/2+C/2)
=1/tan(B/2+C/2)
=1/{[tan(B/2)+tan(C/2)]/[1-tan(B/2)tan(C/2)]}
=[1-tan(B/2)tan(C/2)]/[tan(B/2)+tan(C/2)]
所以左边...
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A/2=(180-B-C)/2=90-(B+C)/2
所以tan(A/2)=tan[90-(B+C)/2]
=cot(B/2+C/2)
=1/tan(B/2+C/2)
=1/{[tan(B/2)+tan(C/2)]/[1-tan(B/2)tan(C/2)]}
=[1-tan(B/2)tan(C/2)]/[tan(B/2)+tan(C/2)]
所以左边=tan(A/2)[tan(B/2)+tan(C/2)]+tan(B/2)tan(C/2)
=[1-tan(B/2)tan(C/2)]/[tan(B/2)+tan(C/2)]*[tan(B/2)+tan(C/2)]+tan(B/2)tan(C/2)
=1-tan(B/2)tan(C/2)+tan(B/2)tan(C/2)
=1=右边
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