11题,谢谢!!!
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A/2+B/2+C/2=90°
A/2=90°-(B/2+C/2)
tanA/2 = tan(90°-(B/2+C/2))
= cot(B/2+C/2)=1/tan(B/2+C/2)
=(1-tanB/2tanC/2)/(tanB/2+tanC/2)
所以tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2
= tanA/2 (tanB/2+tanC/2) +tanB/2*tanC/2
= (1-tanB/2tanC/2)/(tanB/2+tanC/扒喊2)*(tanB/2+tanC/2) +tanB/2*tanC/2
= 1-tanB/2tanC/2+tanB/2*tanC/2=1
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A/2=90°-(B/2+C/2)
tanA/2 = tan(90°-(B/2+C/2))
= cot(B/2+C/2)=1/tan(B/2+C/2)
=(1-tanB/2tanC/2)/(tanB/2+tanC/2)
所以tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2
= tanA/2 (tanB/2+tanC/2) +tanB/2*tanC/2
= (1-tanB/2tanC/2)/(tanB/2+tanC/扒喊2)*(tanB/2+tanC/2) +tanB/2*tanC/2
= 1-tanB/2tanC/2+tanB/2*tanC/2=1
如果你认可我的回答,请点击左下角的“采纳为满意答段虚案”,祝学习进步!
手机提问的朋友在客户端右上角评价握此燃点【满意】即可
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2014-04-13
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