IfA and B are acute angles and $$\sec A = 3; \cot B = 4$$, then the value of $$\frac{\cosec^2 A + \sin^2 B}{\cot^2 A + \sec^2 B}$$ is:

Given,  $$\sec A = 3$$

$$=$$>  $$\cos A=\frac{1}{3}$$

$$=$$>  $$\sin A=\sqrt{1-\cos^2A}=\sqrt{1-\left(\frac{1}{3}\right)^2}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$$

$$=$$>  $$\operatorname{cosec}A=\frac{3}{2\sqrt{2}}$$

$$=$$>  $$\cot A=\frac{\cos A}{\sin A}=\frac{\frac{1}{3}}{\frac{2\sqrt{2}}{3}}=\frac{1}{2\sqrt{2}}$$

$$\cot B=4$$

$$=$$>  $$\tan B=\frac{1}{4}$$

$$=$$>  $$\sec B=\sqrt{1+\tan^2B}=\sqrt{1+\left(\frac{1}{4}\right)^2}=\sqrt{1+\frac{1}{16}}=\sqrt{\frac{17}{16}}=\frac{\sqrt{17}}{4}$$

$$=$$>  $$\cos B=\frac{4}{\sqrt{17}}$$

$$=$$>  $$\sin B=\sqrt{1-\cos^2B}=\sqrt{1-\frac{16}{17}}=\sqrt{\frac{1}{17}}=\frac{1}{\sqrt{17}}$$

$$\therefore\ $$ $$\frac{\operatorname{cosec}^2A+\sin^2B}{\cot^2A+\sec^2B}=\frac{\left(\frac{3}{2\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{17}}\right)^2}{\left(\frac{1}{2\sqrt{2}}\right)^2+\left(\frac{\sqrt{17}}{4}\right)^2}$$

$$=\frac{\frac{9}{8}+\frac{1}{17}}{\frac{1}{8}+\frac{17}{16}}$$

$$=\frac{\frac{153+8}{8\times17}}{\frac{2+17}{16}}$$

$$=\frac{161}{8\times17}\times\frac{16}{19}$$

$$=\frac{322}{323}$$

Hence, the correct answer is Option B