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

$$\frac{1}{\operatorname{cosec}\theta-1}+\frac{1}{\operatorname{cosec}\theta+1}=2\sec\theta$$

$$\frac{\operatorname{cosec}\theta\ +1+\operatorname{cosec}\theta\ -1}{\operatorname{cosec}^2\theta-1}=2\sec\theta$$

$$\frac{2\operatorname{cosec}\theta\ \ }{\cot^2\theta\ }=2\sec\theta$$

$$\frac{\operatorname{cosec}\theta\ }{\sec\theta\ }\ =\cot^2\theta\ $$

$$\cot\theta\ =\cot^2\theta\ $$

$$\cot^2\theta\ -\cot\theta\ =0$$

$$\cot\theta\ \left(\cot\theta\ -1\right)\ =0$$

$$\Rightarrow$$  $$\cot\theta=0$$  or   $$\cot\theta-1=0$$

$$\Rightarrow$$  $$\cot\theta=0$$  or  $$\cot\theta=1$$

$$\Rightarrow$$  $$\theta=90^{\circ\ }$$  or   $$\theta=45^{\circ\ }$$

Since $$0^\circ < \theta < 90^\circ$$

$$\Rightarrow$$  $$\theta=45^{\circ\ }$$

$$\therefore\ $$ $$\cot\theta+\cos\theta=\cot45^{\circ\ }+\cos45^{\circ\ }$$

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

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

$$=\frac{\left(\sqrt{2}+1\right)\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}$$

$$=\frac{2 + \sqrt2}{2}$$

Hence, the correct answer is Option B