If $$\tan a=\frac{2}{\sqrt{13}}$$,then the value of $$\frac{\operatorname{cosec}^2a+2\sec^2a}{\operatorname{cosec}^2a-3\sec^2a}$$ is

Given,  $$\tan a=\frac{2}{\sqrt{13}}$$

$$\frac{\operatorname{cosec}^2a+2\sec^2a}{\operatorname{cosec}^2a-3\sec^2a}=\frac{\frac{1}{\sin^2a}+\frac{2}{\cos^2a}}{\frac{1}{\sin^2a}-\frac{3}{\cos^2a}}$$

$$=\frac{\cos^2a+2\sin^2a}{\cos^2a-3\sin^2a}$$

$$=\frac{\cos^2a\left(1+\frac{2\sin^2a}{\cos^2a}\right)}{\cos^2a\left(1-\frac{3\sin^2a}{\cos^2a}\right)}$$

$$=\frac{1+2\tan^2a}{1-3\tan^2a}$$

$$=\frac{1+2\left(\frac{2}{\sqrt{13}}\right)^2}{1-3\left(\frac{2}{\sqrt{13}}\right)^2}$$

$$=\frac{1+\frac{8}{13}}{1-\frac{12}{13}}$$

$$=\frac{13+8}{13-12}$$

$$=21$$

Hence, the correct answer is Option D