JEE Advanced 2024 · Paper 1

JEE Advanced 2024 Mathematics question on Trigonometry

Let π2<x<π\dfrac{\pi}{2}<x<\pi be such that cot⁡x=−511\cot x=\dfrac{-5}{\sqrt{11}}. Then (sin⁡11x2) ⁣(sin⁡6x−cos⁡6x)+(cos⁡11x2) ⁣(sin⁡6x+cos⁡6x)\left(\sin\dfrac{11x}{2}\right)\!\left(\sin 6x-\cos 6x\right)+\left(\cos\dfrac{11x}{2}\right)\!\left(\sin 6x+\cos 6x\right) is equal to
  1. A.
    11−123\dfrac{\sqrt{11}-1}{2\sqrt{3}}
  2. B.
    11+123\dfrac{\sqrt{11}+1}{2\sqrt{3}}
  3. C.
    11+132\dfrac{\sqrt{11}+1}{3\sqrt{2}}
  4. D.
    11−132\dfrac{\sqrt{11}-1}{3\sqrt{2}}
Show answer

Correct option: B

[JEE Advanced 2024 · Paper 1]

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