If $$\cot \theta = \frac{3}{\sqrt 5}$$, 0$$^\circ$$ < $$\theta$$ < 90$$^\circ,$$ then the value of $$\frac{6 \sec^2 \theta - \frac{5}{3} cosec^2 \theta}{\frac{3}{5} \sec^2 \theta + \frac{4}{3} cosec^2 \theta}$$, is equal to

Given,  $$\cot \theta = \frac{3}{\sqrt 5}$$

$$\Rightarrow$$  $$\tan\theta=\frac{\sqrt{5}}{3}$$

$$\Rightarrow$$  $$\sec^2\theta\ =1+\tan^2\theta\ =1+\left(\frac{\sqrt{5}}{3}\right)^2=1+\frac{5}{9}=\frac{14}{9}$$

$$\Rightarrow$$  $$\operatorname{cosec}^2\theta\ =1+\cot^2\theta\ =1+\left(\frac{3}{\sqrt{5}}\right)^2=1+\frac{9}{5}=\frac{14}{5}$$

$$\therefore\ $$ $$\frac{6\sec^2\theta-\frac{5}{3}\operatorname{cosec}^2\theta}{\frac{3}{5}\sec^2\theta+\frac{4}{3}\operatorname{cosec}^2\theta}=\frac{6\left(\frac{14}{9}\right)-\frac{5}{3}\left(\frac{14}{5}\right)}{\frac{3}{5}\left(\frac{14}{9}\right)+\frac{4}{3}\left(\frac{14}{5}\right)}$$

$$=\frac{\frac{28}{3}-\frac{14}{3}}{\frac{14}{15}+\frac{56}{15}}$$

$$=\frac{\frac{14}{3}}{\frac{70}{15}}$$

$$=\frac{14}{3}\times\frac{15}{70}$$

$$=1$$

Hence, the correct answer is Option B