Introduction to Trigonometry
RD Sharma
CBSE
Grade 10
Question:
If $\sec \theta + \tan \theta = p$, then $\sec \theta$ is equal to:
(a) $\dfrac{p^2 + 1}{2p}$
(b) $\dfrac{p^2 - 1}{2p}$
(c) $\dfrac{p^2 + 1}{p}$
(d) $\dfrac{p^2 - 1}{p}$
Step-by-Step Solution
Key Concept: $\sec \theta - \tan \theta = 1/p$. Add to $\sec \theta + \tan \theta = p \Rightarrow 2\sec \theta = p + 1/p \Rightarrow \sec \theta = \dfrac{p^2 + 1}{2p}$.
$2\sec\theta = p + \dfrac{1}{p} = \dfrac{p^2 + 1}{p} \Rightarrow \sec\theta = \dfrac{p^2 + 1}{2p}$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Solving $\sec \theta = (p^2 + 1)/(2p)$: 1.0 Mark
Correct Answer: $\dfrac{p^2 + 1}{2p}$
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