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△ABC,恒有tan(A/2)+tan(B/2)tan(A/2)+tan(C/2)=sec(A/2)^2
题目详情
△ABC,恒有【tan(A/2)+tan(B/2)】【tan(A/2)+tan(C/2)】=【sec(A/2)】^2
▼优质解答
答案和解析
证明：【tan(A/2)+tan(B/2)】【tan(A/2)+tan(C/2)】
＝tan²(A/2)+tan(A/2)*[tan(B/2)+tan(C/2)]+tan(B/2)*tan(C/2)
=tan²(A/2)+tan(A/2)*tan(B/2 + C/2)*[1-tan(B/2)*tan(C/2)]+tan(B/2)*tan(C/2)
=tan²(A/2)+tan(A/2)*tan(90°-A/2)*[1-tan(B/2)*tan(C/2)]+tan(B/2)*tan(C/2)
=tan²(A/2)+1
=【sec(A/2)】^2
等式得证
＝tan²(A/2)+tan(A/2)*[tan(B/2)+tan(C/2)]+tan(B/2)*tan(C/2)
=tan²(A/2)+tan(A/2)*tan(B/2 + C/2)*[1-tan(B/2)*tan(C/2)]+tan(B/2)*tan(C/2)
=tan²(A/2)+tan(A/2)*tan(90°-A/2)*[1-tan(B/2)*tan(C/2)]+tan(B/2)*tan(C/2)
=tan²(A/2)+1
=【sec(A/2)】^2
等式得证
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