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$\sin^{2} A + \sin^{2} (A - B) + 2 \sin A \cos B \sin (B - A)$ is equal to

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detailed solution

Correct option is B

sin2⁡A+sin2⁡(A−B)+[sin⁡(A+B)+sin⁡(A−B)]sin⁡(B−A) =sin2⁡A+sin2⁡(A−B)+sin⁡(A+B)sin⁡(B−A)−sin2⁡(A−B) =sin2⁡A+sin2⁡B−sin2⁡A=sin2⁡B

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sin2⁡A+sin2⁡(A−B)+2sin⁡Acos⁡Bsin⁡(B−A) is equal to

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$\sin^{2} A + \sin^{2} (A - B) + 2 \sin A \cos B \sin (B - A)$ is equal to