If sum r=1 13 \ frac 1 sin (π/4+(r-1) π/6 ) sin (π/4+r π/6 ) \ =a √(3)+b , a , b in Z , then a 2 + b 2 is equal to :

Mathematics · Trigonometry Ratios and Identities

JEE Main 2025 — 28 January, Evening Shift — Question 16

If $\sum_{r=1}^{13}\left\{\frac{1}{\sin \left(\frac{\pi}{4}+(r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{r \pi}{6}\right)}\right\}=a \sqrt{3}+b$ , $\mathrm{a}, \mathrm{b} \in \mathbf{Z}$ , then $\mathrm{a}^{2}+\mathrm{b}^{2}$ is equal to :

Option A:
10
Option B:
2
Option C:
8
Correct
Option D:
4

Answer: C

Step-by-step solution

$\frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13} \frac{\sin \left[\left(\frac{\pi}{4}+\frac{r \pi}{6}\right)-\left(\frac{\pi}{4}\right)-(r-1) \frac{\pi}{6}\right]}{\sin \left(\frac{\pi}{4}+(r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{r \pi}{6}\right)}$

$\frac{1}{\sin \frac{\pi}{6}} \sum_{\mathrm{r}=1}^{13}\left(\cot \left(\frac{\pi}{4}+(\mathrm{r}-1) \frac{\pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)\right)$

$=2 \sqrt{3}-2=\alpha \sqrt{3}+b$

So $\mathrm{a}^{2}+\mathrm{b}^{2}=8$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Trigonometry Ratios and Identities
Topic: Continued Sum or Product of Series of Trigonometric Ratios