AP EAMCET · Maths · Trigonometric Ratios & Identities
\(\sum_{k=0}^{12} \frac{1}{\sin \left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right) \sin \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)}=\)
- A \(2(\sqrt{3}+1)\)
- B \(2(3-\sqrt{3})\)
- C \(2(2-\sqrt{3})\)
- D \(2(\sqrt{3}-1)\)
Answer & Solution
Correct Answer
(D) \(2(\sqrt{3}-1)\)
Step-by-step Solution
Detailed explanation
The general term of the sum is \(\frac{1}{\sin A_k \sin B_k}\) where \(A_k = (k+1) \frac{\pi}{6}+\frac{\pi}{4}\) and \(B_k = \frac{k \pi}{6}+\frac{\pi}{4}\). We use the identity \(\frac{1}{\sin A \sin B} = \frac{1}{\sin(A-B)} (\cot B - \cot A)\). \(A_k - B_k = \frac{\pi}{6}\),…
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