If $$\frac{1}{\cosec \theta + 1} + \frac{1}{\cosec \theta -1} = 2 \sec \theta, 0^\circ < \theta < 90^\circ$$, then the value of $$\frac{\tan \theta + 2 \sec \theta}{\cosec \theta}$$ is:

Given that $$\dfrac{1}{\cosec\theta +1} + \dfrac {1}{\cosec \theta -1} = 2 \sec \theta $$

$$\Rightarrow \dfrac {\cosec \theta - 1 + \cosec \theta +1}{ \cosec^2 \theta -1} = 2\sec \theta $$

$$\Rightarrow \dfrac{2 \cosec \theta}{\cot^2 \theta} = 2 \sec \theta $$

$$\Rightarrow \dfrac {1}{\sin \theta} \times \dfrac {\sin^2 \theta}{\cos^2 \theta} = \dfrac {1}{\cos \theta} $$

$$\Rightarrow \dfrac{\sin \theta}{\cos \theta} = 1 $$

$$\Rightarrow  \tan \theta =1$$

$$\Rightarrow \theta = \dfrac{\pi}{4}$$

then value $$\dfrac {\tan \theta + 2 \sec \theta}{\cosec \theta}$$

$$\Rightarrow \dfrac {\tan \dfrac{\pi}{4} + 2 \sec \dfrac {\pi}{4}}{\cosec \dfrac{\pi}{4}}$$  (put the value $$\theta $$)

$$\Rightarrow \dfrac{1 + 2\sqrt{2}} {\sqrt {2}}$$ multiply by $$ \sqrt {2}$$

$$\Rightarrow \dfrac {4 + \sqrt {2}}{2} $$ Ans