Question

1-sin/1+sin=(sec-tan)^2
pls ans ASAP
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Answer 1

$LHS = \frac{(1-sin \theta)}{(1+sin \theta)} \\= \frac{(1-sin \theta)(1-sin \theta)}{(1+sin \theta)(1-sin \theta)}$

$= \frac{(1-sin \theta)^{2}}{1^{2} - sin^{2} \theta} \\= \frac{(1-sin \theta)^{2}}{cos^{2} \theta}$

$= \Big(\frac{(1-sin\theta)}{cos\theta}\Big)^{2}\\= \Big( \frac{1}{cos\theta} - \frac{sin \theta}{cos \theta}\Big)^{2}$

$= \Big( sec\theta - tan \theta\Big)^{2} \\= RHS$

Therefore.,

$\red {\frac{(1-sin \theta)}{(1+sin \theta)}}\green {= \Big( sec\theta - tan \theta\Big)^{2}}$

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