Chapter 11: Problem 5
$$ \sqrt{\frac{1-\sin A}{1+\sin A}}=\sec A-\tan A $$
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Chapter 11: Problem 5
$$ \sqrt{\frac{1-\sin A}{1+\sin A}}=\sec A-\tan A $$
These are the key concepts you need to understand to accurately answer the question.
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$$ \cos A \cos (B-A)-\sin A \sin (B-A)=\cos B $$
$$ \tan \theta \sin \left(\frac{\pi}{2}+\theta\right) \cos \left(\frac{\pi}{2}-\theta\right)=\sin ^{2} \theta $$
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$$ \text { If } \tan \theta=\frac{p}{q}, \text { show that } \frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}=\frac{p^{2}-q^{2}}{p^{2}+q^{2}} \text { . } $$
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