tanB/2+cotB/2=10/3
(sinB/2)/(cosB/2)+(cosB/2)/(sinB/2)=10/3
[(sinB/2)^2+(cosB/2)^2]/[(cosB/2)(sinB/2)]=10/3
1/[(sinB)/2]=10/3
sinB=3/5
因(sinB)^2+(cosB)^2=1所以cosB=4/5,
同理,sinA=12/13
cos(A-B)
=cosA*cosB+sinA*sinB
=(5/13)*(4/5)+(12/13)*(3/5)
=56/65