在三角形ABC中，已知a+b=10，c=8，求tan(A/2)*tan(B/2)的值
热心网友
(a+b)/c=10/8---(sinA+sinB)/sinC=5/4---(sinA+sinB)/sin(A+B)=5/4---2sin[(A+B)/2]cos[(A-B)/2]/{2sin[(A+B)/2]cos[(A+B)/2]}=5/4---cos[(A-B)/2]/cos[(A+B)/2]=5/4---[cos(A/2)cos(B/2)+sin(A/2)sin(B/2)]/[......-......]=5/4---4cos(A/2)cos(B/2)+4sin(A/2)sin(B/2)=5*......-5*......---9sin(A/2)sin(B/2)=cos(A/2)cos(B/2)---tan(A/2)tan(B/2)=1/9.
热心网友
= 1/9