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Prove that : $\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\tan A}{\sec A + 1}$
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LHS = $\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \sqrt{\frac{\frac{1}{\sec A} - 1}{\frac{1}{\sec A} + 1}}$ (1 Mark)
$= \sqrt{\frac{\sec A - 1}{\sec A + 1}}$ (1 Mark)
$= \sqrt{\frac{\sec A - 1}{\sec A + 1} \times \frac{\sec A + 1}{\sec A + 1}} = \sqrt{\frac{\sec^2 A - 1}{(\sec A + 1)^2}} = \frac{\tan A}{\sec A + 1} = \text{RHS}$ (1 Mark)
$= \sqrt{\frac{\sec A - 1}{\sec A + 1}}$ (1 Mark)
$= \sqrt{\frac{\sec A - 1}{\sec A + 1} \times \frac{\sec A + 1}{\sec A + 1}} = \sqrt{\frac{\sec^2 A - 1}{(\sec A + 1)^2}} = \frac{\tan A}{\sec A + 1} = \text{RHS}$ (1 Mark)