Dhairya Shah
Last Activity: 7 Years ago
Given
COS (B-A) = 3/5 ___(1)
TAN A=2COT B
So sinA/cos A = 2 cos B/ sin A
So sinAsinB =2cosAcosB____(2)
We know that cos B-A =cosBcosA + sinBsinA=3/5 from 1
So 3cosAcosB =3/5 from 2
On solving we get cosAcosB= 1/5___(3)
SinAsinB=2/5___(4)
Cos (A+B) = cosAcosB – sinAsinB
= 1/5 – 2/5 from equations 3 and 4
=-1/5