Trigonometry · JEE Advanced previous year questions

2024 1 question

1. Q32024 · 26 May 2024 · Paper 1

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

   1. A.$\frac{\sqrt{11}-1}{2\sqrt{3}}$
   2. B.$\frac{\sqrt{11}+1}{2\sqrt{3}}$
   3. C.$\frac{\sqrt{11}+1}{3\sqrt{2}}$
   4. D.$\frac{\sqrt{11}-1}{3\sqrt{2}}$
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   Answer: B