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JEE Advanced · Mathematics · 8. Trigonometric Equations

For \(0 < \theta < \frac{\pi}{2}\), the solution(s) of \(\sum_{m=1}^6 \operatorname{cosec}\left[\theta+\frac{(m-1) \pi}{4}\right]\) \(\operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2}\) is/are

  1. A \(\frac{\pi}{4}\)
  2. B \(\frac{\pi}{6}\)
  3. C \(\frac{\pi}{12}\)
  4. D \(\frac{5 \pi}{12}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{5 \pi}{12}\)

Step-by-step Solution

Detailed explanation

For \(0 < \theta < \frac{\pi}{2}\),
\(\sum_{m=1}^6 \operatorname{cosec}\left[\theta+\frac{(m-1) \pi}{4}\right] \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right) \)
\( =4 \sqrt{2} \)
\( \Rightarrow \sum_{m=1}^6 \frac{\sin \left[\theta+\frac{m \pi}{4}-\left(\theta+\frac{(m-1) \pi}{4}\right)\right]}{\sin \frac{\pi}{4}\left\{\sin \left(\theta+\frac{(m-1) \pi}{4}\right) \sin \left(\theta+\frac{m \pi}{4}\right)\right\}} \)
\( =4 \sqrt{2}\)
\(\Rightarrow \sum_{m=1}^6 \frac{\cot \left(\theta+\frac{(m-1) \pi}{4}\right)-\cot \left(\theta+\frac{m \pi}{4}\right)}{1 / 2} \)
\( =4 \sqrt{2}\)
\(\Rightarrow \sum_{m=1}^6\left[\cot \left(\theta+\frac{(m-1) \pi}{4}\right)\right. \)
\( \left.-\cot \left(\theta+\frac{m \pi}{4}\right)\right]=4 \)
\( \Rightarrow \cot (\theta)-\cot \left(\theta+\frac{\pi}{4}\right)+\cot \left(\theta+\frac{\pi}{4}\right) \)
\( -\cot \left(\theta+\frac{2 \pi}{4}\right)+\ldots+\cot \left(\theta+\frac{5 \pi}{4}\right) \)
\( -\cot \left(\theta+\frac{6 \pi}{4}\right)=4 \)
\( \Rightarrow \cot \theta-\cot \left(\frac{3 \pi}{2}+\theta\right)=4 \)
\( \Rightarrow \cot \theta+\tan \theta=4 \)
\( \Rightarrow \tan ^2 \theta-4 \tan \theta+1=0 \)
\( \Rightarrow (\tan \theta-2)^2-3=0 \)
\( \Rightarrow(\tan \theta-2+\sqrt{3})(\tan \theta-2-\sqrt{3})=0 \)
\( \Rightarrow \tan \theta=2-\sqrt{3} \text { or } \)
\( \tan \theta=2+\sqrt{3} \)
\( \Rightarrow \theta=\frac{\pi}{12} ; \theta=\frac{5 \pi}{12} \Rightarrow \theta \in\left(0, \frac{\pi}{2}\right)\)
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