If ∑_ r=1 ^ 13 \ (1)/( ((π)/(4)+(r-1) (π)/(6)) ((π)/(4)+(r π)/(6)))\ =a 3 +b, a, b ∈ Z , then a^2+b^2 is equal to :

Correct answer: D. 8 (1)/( (π)/(6)) ∑_ r=1 ^ 13 ( [((π)/(4)+(r π)/(6))-((π)/(4))-(r-1) (π)/(6)])/( ((π)/(4)+(r-1) (π)/(6)) ((π)/(4)+(r π)/(6))) aligned & (1)/( (π)/(6)) ∑_ r=1 ^ 13 ( ((π)/(4)+(r-1) (π)/(6))- ((π)/(4)+ r π 6 )) & =2 3 -2=α 3 +b aligned So a^2+b^2=8

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Trigonometric Ratios & Identities

If ∑_ r=1 ^ 13 \ (1)/( ((π)/(4)+(r-1) (π)/(6)) ((π)/(4)+(r...

\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 :

Options

A $10$
B $4$
C $2$
D $8$

Correct answer

D. $8$

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)}

\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}

So

a^2+b^2=8$

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