If $$2 \sin \theta - 8 \cos^2 \theta + 5 = 0, 0^\circ < \theta < 90^\circ$$, then what is the value of $$(\tan 2\theta + \cosec 2\theta)?$$

$$2 \sin \theta - 8 \cos^2 \theta + 5 = 0$$

$$\Rightarrow2 \sin \theta - 8(1 - \sin^2 \theta) + 5 = 0$$

$$\Rightarrow8 \sin^2 \theta + 2 \sin \theta - 3 = 0$$

$$\Rightarrow8 \sin^2 \theta + 6 \sin \theta- 4 \sin \theta- 3 = 0$$

$$\Rightarrow4\sin \theta(2\sin \theta - 1) + 3(2\sin \theta - 1) = 0$$

$$\Rightarrow(4\sin \theta + 3)(2\sin \theta - 1) = 0$$

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

$$\Rightarrow\theta = 30\degree$$

$$(\tan 2\theta + \cosec 2\theta)$$

= $$(\tan 2 \times 30\degree+ \cosec 2 \times 30\degree)$$

= $$(\tan 60\degree+ \cosec 60\degree)$$

=$$\sqrt3 + \frac{2}{\sqrt3}$$

=$$ \frac{5}{\sqrt3} = \frac{5\sqrt3}{3}$$