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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Mathematics
Chapter
Sequences and Series

JEE Advanced 2016 · Paper 2 · Mathematics · Sequences and Series

The value of ∑ _ k = 1 ^ 13 (1)/( ((π)/(4) + ((k - 1)π)/(6))...

The value of ∑k=1131sin⁡(π4+(k−1)π6)sin⁡(π4+kπ6)\mathop{∑}_{k = 1}^{13}\dfrac{1}{\sin \left(\dfrac{\pi}{4} + \dfrac{\left(k - 1\right)\pi}{6}\right)\sin \left(\dfrac{\pi}{4} + \dfrac{k\pi}{6}\right)} is equal to

Options

  1. A3−33 - \sqrt{3}
  2. B2(3−3)2\left(3 - \sqrt{3}\right)
  3. C2(3−1)2\left(\sqrt{3} - 1\right)
  4. D2(2+3)2\left(2 + \sqrt{3}\right)

Correct answer

C. 2(3−1)2\left(\sqrt{3} - 1\right)

Step-by-step solution

Let π4+(K−1)π6=θ\dfrac{\pi}{4} + ( K - 1 )\dfrac{\pi}{6} = \theta
⇒∑k=1131sinθsin(θ+π6)=1sinπ6∑k=113sinπ6sinθsin(θ+π6)⇒ \mathop{∑}_{k = 1}^{13} \dfrac{1}{sin\theta sin( \theta + \dfrac{\pi}{6} )} = \dfrac{1}{sin\dfrac{\pi}{6}} \mathop{∑}_{k = 1}^{13} \dfrac{sin\dfrac{\pi}{6}}{sin\theta sin( \theta + \dfrac{\pi}{6} )}
=2∑k=113sin⁡(θ+π6−θ)sinθsin(θ+π6)=2∑k=113sin⁡(θ+π6)cosθ−cos⁡(θ+π6)sinθsinθsin(θ+π6)= 2 \mathop{∑}_{k = 1}^{13} \dfrac{\sin ( \theta + \dfrac{\pi}{6} - \theta )}{sin\theta sin( \theta + \dfrac{\pi}{6} )} = 2 \mathop{∑}_{k = 1}^{13} \dfrac{\sin ( \theta + \dfrac{\pi}{6} )cos\theta - \cos ( \theta + \dfrac{\pi}{6} )sin\theta}{sin\theta sin( \theta + \dfrac{\pi}{6} )}
=2∑k=113(cotθ−cot⁡(θ+π6))=2∑k=113(cot⁡(π4+(k−1)π6)−cot⁡(π4+kπ6))= 2 \mathop{∑}_{k = 1}^{13} ( cot\theta - \cot ( \theta + \dfrac{\pi}{6} ) ) = 2 \mathop{∑}_{k = 1}^{13} ( \cot ( \dfrac{\pi}{4} + ( k - 1 )\dfrac{\pi}{6} ) - \cot ( \dfrac{\pi}{4} + \dfrac{k\pi}{6} ) )
=2(cotπ4−cot⁡(π4+π6)+cot⁡(π4+π6)−cot⁡(π4+2π6)+−−−−−cot⁡(π4+12π6)−cot⁡(π4+13π6))= 2( cot\dfrac{\pi}{4} - \cot ( \dfrac{\pi}{4} + \dfrac{\pi}{6} ) + \cot ( \dfrac{\pi}{4} + \dfrac{\pi}{6} ) - \cot ( \dfrac{\pi}{4} + \dfrac{2\pi}{6} ) + - - - - - \cot ( \dfrac{\pi}{4} + \dfrac{12\pi}{6} ) - \cot ( \dfrac{\pi}{4} + \dfrac{13\pi}{6} ) )
=2[cotπ4−cot⁡(π4+13π6)]=2[1−cot29π12]=2[1−cot75ο]=2[1−(2−3)]=2(3−1)= 2[ cot\dfrac{\pi}{4} - \cot ( \dfrac{\pi}{4} + \dfrac{13\pi}{6} ) ] = 2[ 1 - cot\dfrac{29\pi}{12} ] = 2[ 1 - cot{75}^{ο} ] = 2[ 1 - ( 2 - \sqrt{3} ) ] = 2( \sqrt{3 - 1} )
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