If $a c o s^{2} 3 a + b c o s^{4} a = 16 c o s^{6} a + 9 c o s^{2} a$ is an identity, then

a = 1, b = 24

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a = 3, b = 23

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a = 4, b = 24

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a = 2, b = 23

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Solution

The correct option is A $a = 1, b = 24$
$L H S = a c o s^{2} 3 a + b c o s^{4} a = a [4 c o s^{3} a - 3 c o s a]^{2} + b c o s^{4} a = a [16 c o s^{6} a + 9 c o s^{2} a - 24 c o s^{4} a] + b c o s^{4} a = 16 a c o s^{6} a + 9 a c o s^{2} a + (b - 24 a) c o s^{4} a B y c o m p a r i s o n, 16 a = 16 ∴ a = 1 t h e n (b - 24 a) = 0 ∴ 24$

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If $a = 2^{3} \times 3^{2} \times 5$ and $b = 2^{4} \times 3 \times 7^{2}$ , then which of the following is true?

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