If $$11 \sin^{2} \theta - \cos^{2} \theta + 4 \sin \theta - 4 = 0, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\frac{\cos 2\theta + \cot 2 \theta}{\sec 2 \theta - \tan 2 \theta}$$

$$11 \sin^{2} \theta - \cos^{2} \theta + 4 \sin \theta - 4 = 0$$

$$11 \sin^{2} \theta - (1 - \sin^{2} \theta) + 4 \sin \theta - 4 = 0$$

$$12 \sin^{2} \theta + 4 \sin \theta - 5 = 0$$

$$12 \sin^{2} \theta + 10 \sin \theta - 6\sin \theta- 5 = 0$$

$$2\sin \theta(6 \sin\theta + 5) -1(6 \sin\theta + 5) = 0$$

$$(2\sin \theta - 1)(6 \sin\theta + 5) = 0$$

For $$ 0^\circ < \theta < 90^\circ$$,

$$\sin \theta = 1/2$$

$$\theta = 30\degree$$

$$\frac{\cos 2\theta + \cot 2 \theta}{\sec 2 \theta - \tan 2 \theta}$$

On putting the value of $$\theta$$,

$$\frac{\cos 2\times 30 + \cot 2 \times 30}{\sec 2 \times 30 - \tan 2 \times 30}$$

$$\frac{\cos 60 + \cot 60}{\sec60 - \tan 60}$$

$$\frac{\frac{1}{2} + \frac{1}{\sqrt3}}{2 - \sqrt3}$$

$$\frac{2 + \sqrt3}{2\sqrt3(2 - \sqrt3)}$$

$$\frac{2 + \sqrt3}{4\sqrt3- 6}$$

$$\frac{2 + \sqrt3}{4\sqrt3- 6} \times \frac{4\sqrt3 + 6}{4\sqrt3 + 6}$$

$$\frac{(2 + \sqrt3)(4\sqrt3 + 6)}{(4\sqrt3)^2- 6^2} $$

$$\frac{8\sqrt3 + 12 + 12 + 6\sqrt3}{12} $$

$$\frac{12 + 7\sqrt3}{6} $$