The value of ${sin}^{2} B - {sin}^{2} A - {sin}^{2} (A - B) + 2 sin A cos B sin (A - B) =$

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Solution

${sin}^{2} B - {sin}^{2} A - {sin}^{2} (A - B) + 2 sin A cos B sin (A - B)$
$= {sin}^{2} B - {sin}^{2} A - {sin}^{2} (A - B) + [sin (A + B) + sin (A - B)] sin (A - B)]$
$= {sin}^{2} B - {sin}^{2} A - {sin}^{2} (A - B) + sin (A + B) sin (A - B) + {sin}^{2} (A - B)$
$= {sin}^{2} B - {sin}^{2} A + \frac{1}{2} \times [cos 2 B - cos 2 A]$
$= {sin}^{2} B - {sin}^{2} A + \frac{1}{2} [1 - 2 {sin}^{2} B - 1 + 2 {sin}^{2} A]$
$= {sin}^{2} B - {sin}^{2} A - {sin}^{2} B + {sin}^{2} A$
$= 0$

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