Trigonometrical Ratios and Identities 1 Question 19
19. For $0<\theta<\frac{\pi}{2}$, the solution(s) of
$\sum _{m=1}^{6} \operatorname{cosec} \theta+\frac{(m-1) \pi}{4} \operatorname{cosec} \theta+\frac{m \pi}{4}=4 \sqrt{2}$ is/are
(a) $\frac{\pi}{4}$
(b) $\frac{\pi}{6}$
(c) $\frac{\pi}{12}$
(d) $\frac{5 \pi}{12}$
$(2009)$
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Answer:
Correct Answer: 19. (a, b)
Solution:
- For $0<\theta<\frac{\pi}{2}$
$$ \begin{aligned} & \sum _{m=1}^{6} \operatorname{cosec} \theta+\frac{(m-1) \pi}{4} \operatorname{cosec} \theta+\frac{m \pi}{4}=4 \sqrt{2} \\ & \Rightarrow \quad \sum _{m=1}^{6} \frac{1}{\sin \theta+\frac{(m-1) \pi}{4} \sin \theta+\frac{m \pi}{4}}=4 \sqrt{2} \\ & \Rightarrow \sum _{m=1}^{6} \frac{\sin \theta+\frac{m \pi}{4}-\theta+\frac{(m-1) \pi}{4}}{\sin \frac{\pi}{4} \sin \theta+\frac{(m-1) \pi}{4} \sin \theta+\frac{m \pi}{4}}=4 \sqrt{2} \end{aligned} $$
$$ \begin{array}{cc} \Rightarrow & \sum _{m=1}^{6} \frac{\cot \theta+\frac{(m-1) \pi}{4}-\cot \theta+\frac{m \pi}{4}}{1 / \sqrt{2}}=4 \sqrt{2} \\ \Rightarrow \quad & \sum _{m=1}^{6} \cot \theta+\frac{(m-1) \pi}{4}-\cot \theta+\frac{m \pi}{4}=4 \\ \Rightarrow & \cot (\theta)-\cot \theta+\frac{\pi}{4}+\cot \theta+\frac{\pi}{4}-\cot \theta+\frac{2 \pi}{4} \\ & +\ldots+\cot \theta+\frac{5 \pi}{4}-\cot \theta+\frac{6 \pi}{4}=4 \\ \Rightarrow & \cot \theta-\cot \frac{3 \pi}{2}+\theta=4 \\ \Rightarrow & \cot \theta+\tan \theta=4 \\ \Rightarrow & \tan { }^{2} \theta-4 \tan \theta+1=0 \\ \Rightarrow & (\tan \theta-2+\sqrt{3})(\tan \theta-2)^{2}-3=0 \\ \Rightarrow & \tan \theta=2-\sqrt{3} \text { or } \tan \theta=2+\sqrt{3} \\ \Rightarrow & \theta=\frac{\pi}{12} ; \theta=\frac{5 \pi}{12} \end{array} $$
