JEE Mains · Maths · STD 11 - Trigonometrical equations
Let \(S={\theta \in\left(0, \frac{\pi}{2}\right): \sum_{m=1}^{9}}\) \(\sec \left(\theta+(m-1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{m \pi}{6}\right)=-\frac{8}{\sqrt{3}}\) Then.
- A \(S =\left\{\frac{\pi}{12}\right\}\)
- B \(S =\left\{\frac{2 \pi}{3}\right\}\)
- C \(\sum_{\theta \in S} \theta=\frac{\pi}{2}\)
- D \(\sum_{\theta \in S} \theta=\frac{3 \pi}{4}\)
Answer & Solution
Correct Answer
(C) \(\sum_{\theta \in S} \theta=\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
Let \(\alpha=\theta+(m-1) \frac{\pi}{6}\) \(\beta=\theta+m \frac{\pi}{6}\) So, \(\beta-\alpha=\frac{\pi}{6}\) Here,\(\sum_{m=1}^{9} \sec \alpha \cdot \sec \beta=\sum_{m=1}^{9} \frac{1}{\cos \alpha \cdot \cos \beta}\)…
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