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Prove the following trigonometric identity :
$\sqrt{\frac{\text{cosec } A - 1}{\text{cosec } A + 1}} = \sec A - \tan A$
$\sqrt{\frac{\text{cosec } A - 1}{\text{cosec } A + 1}} = \sec A - \tan A$
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Solution : LHS
$= \sqrt{\frac{\text{cosec } A - 1}{\text{cosec } A + 1} \times \frac{\text{cosec } A - 1}{\text{cosec } A - 1}}$
$= \sqrt{\frac{(\text{cosec } A - 1)^2}{\cot^2 A}} = \frac{\text{cosec } A - 1}{\cot A}$
$= \frac{\text{cosec } A}{\cot A} - \frac{1}{\cot A} = \sec A - \tan A = RHS$
$= \sqrt{\frac{\text{cosec } A - 1}{\text{cosec } A + 1} \times \frac{\text{cosec } A - 1}{\text{cosec } A - 1}}$
$= \sqrt{\frac{(\text{cosec } A - 1)^2}{\cot^2 A}} = \frac{\text{cosec } A - 1}{\cot A}$
$= \frac{\text{cosec } A}{\cot A} - \frac{1}{\cot A} = \sec A - \tan A = RHS$