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Trigonometric Ratios and Identities

Section: Mathematics

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Question : 22 of 50

Marks: +1, -0

\sum\limits_{k=0}^{12} \frac{1}{\sin\left((k+1)\frac{\pi}{6}+\frac{\pi}{4}\right) \sin\left(\frac{k\pi}{6}+\frac{\pi}{4}\right)} =

[AP EAPCET 21 May 2025 S1]

$2(\sqrt{3}+1)$
$2(3-\sqrt{3})$
$2(2-\sqrt{3})$
$2(\sqrt{3}-1)$

Solution:

\sum\limits_{k=0}^{12} \frac{1}{\sin\left((k+1)\frac{\pi}{6}+\frac{\pi}{4}\right) \cdot \sin\left(\frac{k\pi}{6}+\frac{\pi}{4}\right)}

Multiplying

N^r

and

D^r

by

\sin\frac{\pi}{6}

,

\frac{1}{\sin\frac{\pi}{6}} \sum\limits_{k=0}^{12} \frac{ \sin\left\{ \left((k+1)\frac{\pi}{6}+\frac{4}{\pi}\right) - \left(\frac{k\pi}{6}+\frac{\pi}{4}\right) \right\} }{ 2 \sum\limits_{k=0}^{12} \left( \left((k+1)\frac{\pi}{6}+\frac{\pi}{4}\right) \cdot \sin\left(\frac{k\pi}{6}+\frac{\pi}{4}\right) \right) }

=2\left[\cot\left(\frac{k\pi}{6}+\frac{\pi}{4}\right) - \cot\left((k+1)\frac{\pi}{6}+\frac{\pi}{4}\right)\right\}

\;\;\; + \left\{ \cot\left(\frac{\pi}{6}+\frac{\pi}{4}\right) - \cot\left(\frac{\pi}{6}+\frac{\pi}{4}\right) \right\}

\;\;\; + \dots + \left\{ \cot\left(\frac{12\pi}{6}+\frac{\pi}{4}\right) - \cot\left(\frac{13\pi}{6}+\frac{\pi}{4}\right) \right\}

=2\left[\cot\frac{\pi}{4} + \cot\left(\frac{\pi}{6}+\frac{\pi}{4}\right)\right]

=2\left[1 - \frac{\sqrt{3}-1}{\sqrt{3}+1}\right] =2\left(\frac{2}{\sqrt{3}+1}\right) =\frac{4(\sqrt{3}-1)}{2}

=2(\sqrt{3}-1)

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