已知在△ABC中,tan A/2 +cot A/2=10/3,cosB=5/13 求cos (A-B)/2

tan A/2 +cot A/2=10/3,
sin(A/2)/cos(A/2)+cos(A/2)/sin(A/2)=10/3,
3*[sin^2(A/2)+cos^2(A/2)]=10*[sin(A/2)*cos(A/2)],
3=5sinA,
sinA=3/5.
分二种情况,A为锐角和钝角.
1)A为锐角时,sinA=3/5,cosB=5/13 ,
cosA=√(1-sin^2A)=4/5,
sinB=√(1-cos^2)=12/13,
cos(A-B)=cosA*cosB+sinA*ainB=56/65,
cos(A-B)/2=√[cos(A-B)+1]=11√65/65.
2)A为钝角时,sinA=3/5,cosB=5/13 ,
cosA=-√(1-sin^2A)=-4/5,
cos(A-B)=cosA*cosB+sinA*ainB=16/65,
cos(A-B)/2=√[cos(A-B)+1]=9√65/65.
所以,
cos (A-B)/2=11√65/65,或9√65/65.