If $$\sec A = \frac{\sqrt{11}}{3}$$, then the value of $$\frac{\cosec^2 A + \tan ^2 A}{\sin^2 A + \cot^2 A}$$ is:

Given,  $$\sec A = \frac{\sqrt{11}}{3}$$

$$=$$>  $$\cos A=\frac{3}{\sqrt{11}}$$

$$\frac{\operatorname{cosec}^2A+\tan^2A}{\sin^2A+\cot^2A}=\frac{\frac{1}{\sin^2A}+\tan^2A}{\sin^2A+\frac{1}{\tan^2A}}$$

$$=\frac{\frac{1}{1-\cos^2A}+\sec^2A-1}{1-\cos^2A+\frac{1}{\sec^2A-1}}$$

$$=\frac{\frac{1}{1-\left(\frac{3}{\sqrt{11}}\right)^2}+\left(\frac{\sqrt{11}}{3}\right)^2-1}{1-\left(\frac{3}{\sqrt{11}}\right)^2+\frac{1}{\left(\frac{\sqrt{11}}{3}\right)^2-1}}$$

$$=\frac{\frac{1}{1-\frac{9}{11}}+\frac{11}{9}-1}{1-\frac{9}{11}+\frac{1}{\frac{11}{9}-1}}$$

$$=\frac{\frac{11}{2}+\frac{2}{9}}{\frac{2}{11}+\frac{9}{2}}$$

$$=\frac{\frac{99+4}{18}}{\frac{4+99}{22}}$$

$$=\frac{103}{18}\times\frac{22}{103}\ $$

$$=\frac{11}{9}$$

Hence, the correct answer is Option B