If sin a - sinB = a and cosa + cosB = b, then
tan(a - b) =
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Let SinA + SinB = a ------------(1)
and CosA + CosB = b ------------(2)
Sq the two eqns, and then subtract them
You will get, Cos2A+Cos2B+2Cos(A+B) = b2-a2.
So, 2Cos(A+B)Cos(A-B)+2Cos(A+B) = b2-a2.
or Cos(A+B){2Cos(A-B)+2} = b2-a2 ----------------(3)
Now Sq the two eqns and add, ie (2)^2+(1)^2
You will get 2+2Cos(A-B) = a2+b2 -----------[Use this in (3)]
You'd get Cos(A+B){a2+b2) = (b2-a2).
or Cos(A+B) = (b2-a2)/{a2+b2)
Step-by-step explanation:
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