$\frac{{sin}^{2} A - {sin}^{2} B}{sin A cos A - sin B cos B} = tan (A - B)$

True

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False

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Solution

The correct option is B False
$\frac{{sin}^{2} A - {sin}^{2} B}{sin A cos A - sin B cos B} = \frac{2 (sin (A + B) sin (A - B))}{2 sin A cos A - 2 sin B cos B}$

$= \frac{2 sin (A + B) sin (A - B)}{sin 2 A - sin 2 B}$

$= \frac{2 sin (A + B) sin (A - B)}{2 cos (A + B) sin (A - B)}$

$= tan (A + B)$

So the given relation is $False$

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sin2A−sin2BsinAcosA−sinBcosB=tan(A−B)

The correct option is B Falsesin2A−sin2BsinAcosA−sinBcosB=2(sin(A+B)sin(A−B))2sinAcosA−2sinBcosB =2sin(A+B)sin(A−B)sin2A−sin2B =2sin(A+B)sin(A−B)2cos(A+B)sin(A−B) =tan(A+B)So the given relation is False

$\frac{{sin}^{2} A - {sin}^{2} B}{sin A cos A - sin B cos B} = tan (A - B)$