Let be such that Cot . Then is equal to :

Engineering

Mathematics

Trigonometric Ratios of Allied Angles

Question

Let $\frac{π}{2} < x < π$ be such that Cot $x = \frac{- 5}{\sqrt{11}}$ . 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 :

$\frac{\sqrt{11} - 1}{2 \sqrt{3}}$

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

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

$\frac{\sqrt{11} - 1}{3 \sqrt{2}}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$x \in (\frac{π}{2}, π)$ $\cot x = - \frac{5}{\sqrt{11}}$

$(\sin \frac{11 x}{2}) (\sin 6 x - \cos 6 x) + (\cos \frac{11 x}{2}) (\sin 6 x + \cos 6 x)$ $\frac{1 - \tan^{2} \frac{x}{2}}{2 \tan \frac{x}{2}} = - \frac{5}{\sqrt{11}}$

$= \{\sin 6 xsin \frac{11 x}{2} + \cos \frac{11 x}{2} \cos 6 x\}$ $\tan \frac{x}{2} = \sqrt{11}, - \frac{1}{\sqrt{11}}$

$= \cos (6 x - \frac{11 x}{2}) + \sin (6 x - \frac{11 x}{2})$ $\tan \frac{x}{2} = \sqrt{11}, As \frac{π}{4} < \frac{x}{2} < \frac{π}{2}$

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

$= \frac{1}{2 \sqrt{3}} + \frac{\sqrt{11}}{2 \sqrt{3}}$

$= \frac{\sqrt{11} + 1}{2 \sqrt{3}} \Rightarrow$ Option (B) is correct.