$\frac{sin A - 2 {sin}^{3} A}{2 {cos}^{3} A - cos A} =$

$sec A$

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$cot A$

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$tan A$

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$1$

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Solution

The correct option is C
$tan A$

$\frac{sin A - 2 {sin}^{3} A}{2 {cos}^{3} A - c o s A} = \frac{sin A (1 - 2 {sin}^{2} A)}{cos A (2 {cos}^{2} A - 1)}$

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

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

$= tan A$

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