Question

prove
$\sqrt{1 - \sin( \alpha ) } \div \sqrt{1 + \sin( \alpha ) } = \sec( \alpha ) - \tan( \alpha )$
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Answer 1

Answer:

$\sqrt{\frac{1+sin\theta}{1-sin\theta}}+\sqrt{\frac{1-sin\theta}{1+sin\theta}}=2sec\thetaStep-by-step explanation:\sqrt{\frac{1+sin\theta}{1-sin\theta}}+\sqrt{\frac{1-sin\theta}{1+sin\theta}}=\frac{\left(\sqrt{1+sin\theta}\right)^{2}+\left(\sqrt{1-sin\theta}\right)^{2}}{\sqrt{(1-sin\theta)(1+sin\theta)}}=\frac{1-sin\theta+1+sin\theta}{\sqrt{1^{2}-sin^{2}\theta}}=\frac{2}{\sqrt{cos^{2}\theta}}= \frac{2}{cos\theta}= 2sec\theta= RHS Therefore,\sqrt{\frac{1+sin\theta}{1-sin\theta}}+qrt{\frac$

Step-by-step explanation:

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