If $$\frac{1}{1 - \sin \theta} + \frac{1}{1 + \sin \theta} = 4 \sec \theta, 0^\circ < \theta < 90^\circ$$, then the value of $$(3 \cot \theta + \cosec \theta)$$ is:

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

$$\frac{2}{1-\sin^2\theta\ }=\sec\theta\ $$

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

$$2\sec^2\theta\ =\sec\theta\ $$

$$\sec\theta\ =2$$

$$\frac{h}{b}=\frac{2}{1}$$

p=$$\sqrt{4-1}=\sqrt{3}$$

$$(3 \cot \theta + \cosec \theta)=3\times\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}\ =\frac{5}{\sqrt{3}}$$

$$\frac{5}{\sqrt{3}}\times\ \frac{\sqrt{\ 3}}{\sqrt{\ 3}}=\frac{5\sqrt{\ 3}}{3}$$