${cos}^{2} A (3 - 4 {cos}^{2} A)^{2} + {sin}^{2} A (3 - 4 {sin}^{2} A)^{2}$ is equal to-

cos 4 A

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sin 4 A

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1

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None of these

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Solution

The correct option is D $1$
$c o s^{2} A (3 - 4 c o s^{2} A) + s i n^{2} A (3 - 4 s i n^{2} A) = (3 c o s A - 4 c o s^{3} A)^{2} + (3 s i n A - 4 s i n^{3} A)^{2} = (4 c o s^{3} A - 3 c o s A)^{2} + (3 s i n A - 4 s i n^{3} A)^{2} = (c o s 3 A)^{2} + (s i n 3 A)^{2} = 1$

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\frac{{cos}^{4} A}{{cos}^{2} B} + \frac{{sin}^{4} A}{{sin}^{2} B} = 1

\frac{{cos}^{4} B}{{cos}^{2} A} + \frac{{sin}^{4} B}{{sin}^{2} A}

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