Question

1.在△ABC中,tan(A/2)+cot(A/2)=10/3,cosB=5/13,(1)求cos(A-B)的值,(2)

Answer 1

第一题：

(1) tan(A/2)+cot(A/2)=1/(sinA/2*cosA/2)=2/sinA=10/3,=>sinA=3/5;

∵sinC=sin(A+B)>0 => cosA>-1/4 ∴cosA=4/5;cosB=5/13 sinB=12/13; cos(A-B)=cosAcosB+sinAsinB=56/65;

(2) ∵cos(A-B)>0 cosA>cosB ∴A