If $$\sin \theta = \sqrt{3} \cos \theta, 0^\circ < \theta < 90^\circ$$, then the value of $$2 \sin^2 \theta + \sec^2 \theta + \sin \theta \sec \theta + \cosec \theta$$ is:

$$\sin \theta = \sqrt{3} \cos \theta, 0^\circ < \theta < 90^\circ$$
$$\Rightarrow  \frac{\sin \theta}{\cos \theta} = \sqrt{3}$$
$$\Rightarrow \tan \theta = \sqrt{3}$$ 
$$\Rightarrow \theta = 60\degree$$
Now,
$$2 \sin^2 \theta + \sec^2 \theta + \sin \theta \sec \theta + \cosec \theta$$
= $$2 \sin^2 60\degree + \sec^2 60\degree + \tan 60\degree + \cosec60\degree$$
= $$2 \times (\frac{\sqrt{3}}{2})^2 + 2^2 + \sqrt{3} + \frac{2}{\sqrt{3}}$$
= $$\frac{3}{2} + 4 + \sqrt{3} + \frac{2}{\sqrt{3}}$$
= $$\frac{3\sqrt{3} + 8\sqrt{3} + 6 + 4}{2\sqrt{3}}$$
= $$\frac{11\sqrt{3} + 10}{2\sqrt{3}}$$
= $$\frac{33 + 10\sqrt{3}}{6}$$
$$\therefore$$ The correct answer is option A.