已知cosA=5/13,c=21,tanB/2+cotB/2=10/3,求cos(A-B)的值及三角形ABC的面积

设A,B的对边为a,b
tanB/2+cotB/2=10/3,通分花简得sinB=3/5
A,B为三角形内角,得cosB=4/5,sinA=12/13
∴cos(A-B)=56/65
sinC=sin( π-A-B)=sin(A+B)=63/65
由正弦定理得21/sinC=b/sinB=a/sinA
解得a=20,b=13
∴S△ABC=1/2×a×c×sinB=126