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

{sin}^{2} A

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{sin}^{2} B

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{cos}^{2} A

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{cos}^{2} B

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Solution

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

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Similar questions

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

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(i i) (t a n^{2} A s e c^{2} B - s e c^{2} A t a n^{2} B) = (t a n^{2} A - t a n^{2} B)

a s i n^{2} x + b c o s^{2} x = c,

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\frac{a^{2}}{b^{2}}

{cot}^{2} A - {cot}^{2} B = \frac{{cos}^{2} A - {cos}^{2} B}{{sin}^{2} A {sin}^{2} B}

{cos}^{2} 73^{o} + {cos}^{2} 47^{o} - {sin}^{2} 43^{o} + {sin}^{2} 107^{o}

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