Introduction to Trigonometry
RD Sharma
CBSE
Grade 10
Question:
If $\sin \theta + \cos \theta = p$ and $\sec \theta + \csc \theta = q$, prove that $q(p^2 - 1) = 2p$.
Step-by-Step Solution
Key Concept: $p^2 = (\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta \Rightarrow p^2 - 1 = 2 \sin \theta \cos \theta$.<br>$q = \sec \theta + \csc \theta = \dfrac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = \dfrac{p}{\sin \theta \cos \theta}$.<br>$q(p^2 - 1) = \left(\dfrac{p}{\sin \theta \cos \theta}\right)(2 \sin \theta \cos \theta) = 2p$.
$p^2 = 1 + 2\sin\theta\cos\theta \Rightarrow p^2 - 1 = 2\sin\theta\cos\theta$. [1.0 Mark]
$q = \dfrac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} = \dfrac{p}{\sin\theta\cos\theta}$. [1.0 Mark]
$q(p^2 - 1) = \dfrac{p}{\sin\theta\cos\theta} \times 2\sin\theta\cos\theta = 2p$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding $p^2 - 1 = 2\sin\theta\cos\theta$: 1.0 Mark
Expressing $q = p/(\sin\theta\cos\theta)$: 1.0 Mark
Evaluating $q(p^2 - 1) = 2p$: 1.0 Mark
Correct Answer:
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