What is the relation between p and q?
For the following two (02) items: Let \sin \theta + \cos \theta = p and \sec \theta + \operatorname{cosec} \theta = q, where p \neq 1.
- A. p = q(p^2 - 1)
- B. 2p = q(p^2 - 1) ✓
- C. q = p^2 - 1
- D. 2q = p(p^2 - 1)
Correct Answer: B. 2p = q(p^2 - 1)
Explanation
Squaring the first equation gives 1 + 2\sin\theta\cos\theta = p^2, so 2\sin\theta\cos\theta = p^2-1. The second equation is \frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta} = q. Substituting p and \frac{p^2-1}{2} yields \frac{2p}{p^2-1} = q, leading to 2p = q(p^2-1).
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