Mathematics · 2025
JEE Main · 28 January 2025, Shift 2 · Q13
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:
If $\displaystyle \sum_{r=1}^{13}\left\{\frac{1}{\sin \left(\dfrac{\pi}{4}+(r-1) \dfrac{\pi}{6}\right) \sin \left(\dfrac{\pi}{4}+\dfrac{r \pi}{6}\right)}\right\}=a \sqrt{3}+b, a, b \in \mathbf{Z}$, then $\displaystyle a^2+b^2$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 8$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.