If $$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=5$$, then the value of $$\frac{4\sin^2\theta+3}{2\cos^2\theta+2}$$ is:

$$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=5$$

$$\sin\theta+\cos\theta=5\sin\theta\ -5\cos\theta\ $$

$$4\sin\theta=6\cos\theta\ $$

$$\tan\theta\ =\frac{3}{2}$$

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

$$\cos\theta\ =\frac{2}{\sqrt{13}}$$

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

$$\frac{4\sin^2\theta+3}{2\cos^2\theta+2}=\frac{4\left(\frac{3}{\sqrt{13}}\right)^2+3}{2\left(\frac{2}{\sqrt{13}}\right)^2+2}$$

$$=\frac{\frac{36}{13}+3}{\frac{8}{13}+2}$$

$$=\frac{\frac{36+39}{13}}{\frac{8+26}{13}}$$

$$=\frac{75}{34}$$

Hence, the correct answer is Option B