sin6π1r=1∑13sin(4π+(r−1)6π)sin(4π+6rπ)sin[(4π+6rπ)−(4π)−(r−1)6π] $\begin{aligned}
& \frac{1}{\sin \frac{\pi}{6}} \sum_{\mathrm{r}=1}^{13}\left(\cot \left(\frac{\pi}{4}+(\mathrm{r}-1) \frac{\pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)\right) \
& =2 \sqrt{3}-2=\alpha \sqrt{3}+\mathrm{b}
\end{aligned}Soa^2+b^2=8$
If Σ_r=1^13 1 sin ( π 4+(r-1) π 6 ) sin ( π 4+ r π 6 ) =a √3+b, a, b ∈ Z, then a^2+b^2 is equal to :
JEE Main 2025 — Mathematics Trigonometry
If r=1∑13{sin(4π+(r−1)6π)sin(4π+6rπ)1}=a3+b,a,b∈Z, then a2+b2 is equal to :
Held on 28 Jan 2025 · Verified 6 Jul 2026.
Options
- A
10
- B
4
- C
2
- D
8
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