If $$\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}=5$$ and $$\theta$$ is an acute angle, then the value of $$\frac{3\cos^2\theta+1}{3\cos^2\theta-1}$$ is:

Given,  $$\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}=5$$

$$\Rightarrow$$  $$\sec\theta+\tan\theta=5\sec\theta\ -5\tan\theta\ $$

$$\Rightarrow$$  $$6\tan\theta=4\sec\theta$$

$$\Rightarrow$$  $$6\frac{\sin\theta\ }{\cos\theta\ }=4\frac{1}{\cos\theta\ }$$

$$\Rightarrow$$  $$\sin\theta=\frac{2}{3}$$

$$\cos^2\theta\ =1-\sin^2\theta\ =1-\left(\frac{2}{3}\right)^2=\frac{5}{9}$$

$$\therefore\ $$ $$\frac{3\cos^2\theta+1}{3\cos^2\theta-1}=\frac{3\left(\frac{5}{9}\right)+1}{3\left(\frac{5}{9}\right)-1}$$

$$=\frac{\frac{5}{3}+1}{\frac{5}{3}-1}$$

$$=\frac{8}{2}$$

$$=4$$

Hence, the correct answer is Option A