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if sinA+sinB= root3(cos B-cos A) then sin3B+sin3A =
Hello Hemanth ....
sinA + sinB = root(3) cosB - root(3)cosA
Divide by 2 both side
sinA / 2 + sinB / 2 = root(3).cosB/2 - root(3).cosA/2
1/2 . sinA + root(3)/2 .cosA = root(3)/2 . cos(B) - 1/2 . sin(B)
sin(A + 60) = cos(B + 30) => sin(A+60) = sin ( 60 - B )
so finaly A + 60 = 60 - B
so A = -B
Therefore Sin3B + sin3A = 0
I hope i have answered you :)
Regards
Yagya
askiitians_expert
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