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JEE Advanced 2016

JEE Main 2016 Mathematics question on Trigonometry

The value of ∑k=1131sin⁡(π4+(k−1)π6)sin⁡(π4+kπ6)\sum_{k=1}^{13} \dfrac{1}{\sin\left(\dfrac{\pi}{4} + \dfrac{(k-1)\pi}{6}\right)\sin\left(\dfrac{\pi}{4} + \dfrac{k\pi}{6}\right)} is equal to
  1. A.
    3−33 - \sqrt{3}
  2. B.
    2(3−3)2(3 - \sqrt{3})
  3. C.
    2(3−1)2(\sqrt{3} - 1)
  4. D.
    2(2+3)2(2 + \sqrt{3})
Show answer and solution

Correct option: C

Solution

Using the telescoping identity: 1/(sin A · sin B) = (1/sin(B-A)) · [sin(B-A)/(sin A · sin B)] = (1/sin(π/6)) · [sin(B - A)/(sin A · sin B)]

Here B - A = π/6 for each term, so sin(B - A) = 1/2, giving:

1/(sin A · sin B) = 2·[sin(B-A)/(sin A · sin B)] = 2·[cot A - cot B]

Each term = 2[cot(π/4 + (k-1)π/6) - cot(π/4 + kπ/6)]

Summing from k=1 to 13 telescopes to:

2[cot(π/4) - cot(π/4 + 13π/6)]

cot(π/4) = 1

π/4 + 13π/6 = 3π/12 + 26π/12 = 29π/12 = 2π + 5π/12, so cot(29π/12) = cot(5π/12)

cot(5π/12) = cot(75°) = tan(15°) = 2 - √3

Sum = 2[1 - (2 - √3)] = 2[√3 - 1] = 2(√3 - 1)

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