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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).
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