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Mathematics · Trigonometry Ratios and Identities

JEE Advanced 2024 — Paper 1 — Question 3

Let π2<x<π\frac{\pi}{2}<x<\pi be such that cot⁡x=−511\cot x=\frac{-5}{\sqrt{11}}. Then (sin⁡11x2)(sin⁡6x−cos⁡6x)+(cos⁡11x2)(sin⁡6x+cos⁡6x)\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x) is equal to

  1. Option A:

    11−123\frac{\sqrt{11}-1}{2 \sqrt{3}}

  2. Option B:

    11+123\frac{\sqrt{11}+1}{2 \sqrt{3}}

    Correct
  3. Option C:

    11+132\frac{\sqrt{11}+1}{3 \sqrt{2}}

  4. Option D:

    11−132\frac{\sqrt{11}-1}{3 \sqrt{2}}

Answer: B

Step-by-step solution

cot⁡x=−511,x∈(π2,π)⇒cos⁡x=−56\quad \cot \mathrm{x}=\frac{-5}{\sqrt{11}}, \mathrm{x} \in\left(\frac{\pi}{2}, \pi\right) \Rightarrow \cos \mathrm{x}=\frac{-5}{6}

⇒sin⁡x2=1123\Rightarrow \sin \frac{x}{2}=\frac{\sqrt{11}}{2 \sqrt{3}} and cos⁡x2=123\cos \frac{x}{2}=\frac{1}{2 \sqrt{3}}

Now, (sin⁡11x2)(sin⁡6x−cos⁡6x)+(cos⁡11x2)(sin⁡6x+cos⁡6x)\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)

=cos⁡x2+sin⁡x2=11+123=\cos \frac{x}{2}+\sin \frac{x}{2}=\frac{\sqrt{11}+1}{2 \sqrt{3}}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Introduction to Trigonometry