Q. 3 Let \( \frac{\pi}{2}

To solve this problem, we need to use the given information about \( \cot x \) and then apply trigonometric identities to simplify the expression \( \left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x) \). Given: \[ \frac{\pi}{2} < x < \pi \] \[ \cot x = \frac{-5}{\sqrt{11}} \] First, we find \( \tan x \) since \( \cot x = \frac{1}{\tan x} \): \[ \tan x = \frac{1}{\cot x} = \frac{1}{\frac{-5}{\sqrt{11}}} = -\frac{\sqrt{11}}{5} \] Now, we need to find \( \sin x \) and \( \cos x \). Since \( \tan x = \frac{\sin x}{\cos x} \), we can use the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) to find \( \sin x \) and \( \cos x \). Let \( \sin x = a \) and \( \cos x = b \). Then: \[ \tan x = \frac{a}{b} = -\frac{\sqrt{11}}{5} \] \[ a^2 + b^2 = 1 \] From the first equation, we can express \( a \) in terms of \( b \): \[ a = -\frac{\sqrt{11}}{5}b \] Substitute this into the Pythagorean identity: \[ \left(-\frac{\sqrt{11}}{5}b\right)^2 + b^2 = 1 \] \[ \frac{11}{25}b^2 + b^2 = 1 \] \[ \frac{36}{25}b^2 = 1 \] \[ b^2 = \frac{25}{36} \] \[ b = \pm\frac{5}{6} \] Since \( \frac{\pi}{2} < x < \pi \), \( \cos x \) is negative, so: \[ b = -\frac{5}{6} \] Now we can find \( a \): \[ a = -\frac{\sqrt{11}}{5}b = -\frac{\sqrt{11}}{5} \left(-\frac{5}{6}\right) = \frac{\sqrt{11}}{6} \] So, we have: \[ \sin x = \frac{\sqrt{11}}{6} \] \[ \cos x = -\frac{5}{6} \] Now we need to simplify the expression: \[ \left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x) \] We can use the sum-to-product identities to simplify this expression. First, we need to find \( \sin \frac{11 x}{2} \) and \( \cos \frac{11 x}{2} \). We can use the half-angle identities for sine and cosine: \[ \sin \frac{11 x}{2} = \pm\sqrt{\frac{1 - \cos 11 x}{2}} \] \[ \cos \frac{11 x}{2} = \pm\sqrt{\frac{1 + \cos 11 x}{2}} \] Since \( \frac{\pi}{2} < x < \pi \), \( \frac{11 x}{2} \) will be in the second or third quadrant, where sine is positive and cosine is negative. Therefore, we have: \[ \sin \frac{11 x}{2} = \sqrt{\frac{1 - \cos 11 x}{2}} \] \[ \cos \frac{11 x}{2} = -\sqrt{\frac{1 + \cos 11 x}{2}} \] Now we need to find \( \cos 11 x \). We can use the double angle identity for cosine: \[ \cos 11 x = 2\cos^2 5.5x - 1 \] We can further simplify \( \cos 5.5x \) using the half-angle identity: \[ \cos 5.5x = \pm\sqrt{\frac{1 + \cos 11 x}{2}} \] Since \( \frac{\pi}{2} < x < \pi \), \( 5.5x \) will be in the second or third quadrant, where cosine is negative. Therefore, we