if A and B are the solutions of6cos thita+8sin thita=9 then sin(A+B) ?

Hi Rupali,

As A,B are both solutions of this equation,

We have 6CosA + 8SinA = 6CosB + 8SinB

or 6(cosA-cosB) = 8(sinB-sinA)

or 6*2*sin(B-A)/2*sin(B+A)/2 = 8*2Sin(B-A)/2*cos(B+A)/2

or tan(A+B)/2 = 8/6 = 4/3

Now Sin2X = 2tanx/(1+tan^2x)

Henc Sin(A+B) = [ 2*(4/3) ] / [1 + (4/3)^2] = 24/25.

Hope that helps.

Best Regards,

Ashwin (IIT Madras).

Approved