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

(B) $\frac{\sqrt{11}+1}{2 \sqrt{3}}$

(A) $\frac{\sqrt{11}-1}{2 \sqrt{3}}$

(C) $\frac{\sqrt{11}+1}{3 \sqrt{2}}$

(D) $\frac{\sqrt{11}-1}{3 \sqrt{2}}$

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Trigonometric Ratios and Identities

Let

\frac{π}{2} < x < π

be such that

cot x = \frac{- 5}{11}

. Then

(sin \frac{11 x}{2}) (sin 6 x - cos 6 x) + (cos \frac{11 x}{2}) (sin 6 x + cos 6 x)

is equal to :

Answer: (B)

Given

cot x = \frac{- 5}{11}

with

x \in (\frac{π}{2}, π)

.

Since

cot x = \frac{c o s x}{s i n x} = \frac{- 5}{11}

, we have

cos x = \frac{- 5}{6}

and

sin x = \frac{11}{6}

(using

sin^{2} x + cos^{2} x = 1

).

The given expression can be rewritten:

sin 6 x sin \frac{11 x}{2} + cos 6 x cos \frac{11 x}{2} + sin 6 x cos \frac{11 x}{2} - cos 6 x sin \frac{11 x}{2}

= cos (6 x - \frac{11 x}{2}) + sin (6 x - \frac{11 x}{2})

= cos \frac{x}{2} + sin \frac{x}{2}

Using half-angle formulas with the values of

sin x

and

cos x

sin \frac{x}{2} = \frac{1 - cos x}{2} = \frac{1 + \frac{5}{6}}{2} = \frac{11}{2 3}

cos \frac{x}{2} = \frac{1 + cos x}{2} = \frac{1 - \frac{5}{6}}{2} = \frac{1}{2 3}

Therefore:

cos \frac{x}{2} + sin \frac{x}{2} = \frac{1}{2 3} + \frac{11}{2 3} = \frac{11 + 1}{2 3}

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Single correct option