三角形中a+b=10,c=8则tan(A/2)tan(B/2)的值是

由正弦定理(a+b)/c = a/c + b/c = sinA/sinC + sinB/sinC = (sinA + sinB) / sinC = 10/8 = 5/4
由和差化积公式：sinA + sinB = 2 * sin[(A+B)/2] * cos[(A-B)/2]
2 * sin[(A+B)/2] * cos[(A-B)/2] / sinC = 5/4
因为A+B+C=180度，所以sinC = sin(A+B) = 2 * sin[(A+B)/2] * cos[(A+B)/2] 注：倍角公式
所以2 * sin[(A+B)/2] * cos[(A-B)/2] / (2 * sin[(A+B)/2] * cos[(A+B)/2]) = 5/4
约分 cos[(A-B)/2] / cos[(A+B)/2] = 5/4，用和差公式展开：
[ cos(A/2)cos(B/2) + sin(A/2)sin(B/2) ] / [ cos(A/2)cos(B/2) - sin(A/2)sin(B/2)] = 5/4
分子分母同时除以cos(A/2)cos(B/2)可得：
[ 1 + tan(A/2)tan(B/2) ] / [1 - tan(A/2)tan(B/2) ] = 5/4
解此方程可得
tan(A/2)tan(B/2) = 1/9

a+b=10,c=84(a+b)=5c由正弦定理得4(sinA+sinB)=5sinC=5sin(A+B)8sin(A+B)/2*cos(A-B)/2=10sin(A+B)/2*cos(A+B)/24cos(A-B)/2=5cos(A+B)/24cos(A/2)cos(B/2)+4sin(A/2)sin(B/2)=5cos(A/2)cos(B/2)-5sin(A/2)sin(B/2)9...