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JEE Main 28-Jan-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 28-Jan-2025 Paper - Shift 24 Marks
MCQ
If $\sum_{r=1}^{13}\left\{\frac{1}{\sin \left(\frac{\pi}{4}+(r-1) \frac{\pi}{6}\right) \sin\left(\frac{\pi}{4}+\frac{r \pi}{6}\right)}\right\}=a \sqrt{3}+b$,$a, b \in \mathbf{Z}$, then $a^{2}+b^{2}$ is equal to :
  • A
    10
  • B
    2
  • C
    8
  • D
    4
✓

Answer

C.
$\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_{r=1}^{13}\left(\cot \left(\frac{\pi}{4}+(r-1) \frac{\pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{r \pi}{6}\right)\right)$
$=2 \sqrt{3}-2=\alpha \sqrt{3}+b$
So $\mathrm{a}^{2}+\mathrm{b}^{2}=8$

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