If $$\frac{1}{1 - \sin \theta} + \frac{1}{1 + \sin \theta} = 4 \sec \theta, 0^\circ < \theta < 90^\circ$$, then the value of $$\cot\theta+\operatorname{cosec}\theta$$ is:

$$\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}=4\sec\theta$$

$$\frac{1+\sin\theta\ +1-\sin\theta\ }{1-\sin^2\theta}=\frac{4}{\cos\theta}$$

$$\frac{2}{\cos^2\theta}=\frac{4}{\cos\theta}$$

$$\cos\theta=\frac{1}{2}$$

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

$$\Rightarrow$$  $$\theta=60^{\circ}$$

$$\cot\theta+\operatorname{cosec}\theta=\cot60^{\circ}+\operatorname{cosec}60^{\circ}$$

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

= $$\frac{3}{\sqrt{3}}$$

= $$\sqrt{3}$$

Hence, the correct answer is Option B