Prove that ${sin}^{2} A {cos}^{2} B - {cos}^{2} A {sin}^{2} B = {sin}^{2} A - {sin}^{2} B$

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Solution

LHS $= {(sin A cos B)}^{2} - {(cos A sin B)}^{2}$

$= (sin A cos B + cos A sin B) (sin A cos B - cos A sin B)$

$= sin (A + B) sin (A - B)$

$= {sin}^{2} A - {sin}^{2} B$

$= R H S$

Hence proved

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