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Q.limx01cos4xcos6xsin25x\lim_{x \to 0} \dfrac{1 - \cos 4x \cos 6x}{\sin^2 5x} equals

a
825\dfrac{8}{25}
b
1625\dfrac{16}{25}
c
2425\dfrac{24}{25}
d
3225\dfrac{32}{25}

Correct Answer: Option B

The correct solution involves applying the fundamental concept to derive the final value step by step...

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