If $$\sqrt{3}\cos\theta=\sin\theta$$, then the value of $$\frac{4\sin^2\theta-5\cos\theta}{3\cos\theta+1}$$ is:

Given,  $$\sqrt{3}\cos\theta=\sin\theta$$

$$\Rightarrow$$  $$\sqrt{3}=\frac{\sin\theta}{\cos\theta\ }$$

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

$$\Rightarrow$$  $$\tan\theta\ =\tan60^{\circ\ }$$

$$\Rightarrow$$  $$\theta =60^{\circ\ }$$

$$\therefore\ $$ $$\frac{4\sin^2\theta-5\cos\theta}{3\cos\theta+1}=\frac{4\sin^260^{\circ\ }-5\cos60^{\circ\ }}{3\cos60^{\circ}+1}$$

$$=\frac{4\left(\frac{\sqrt{3}}{2}\right)^2-5\left(\frac{1}{2}\right)}{3\left(\frac{1}{2}\right)+1}$$

$$=\frac{4\left(\frac{3}{4}\right)-\frac{5}{2}}{\frac{3}{2}+1}$$

$$=\frac{3-\frac{5}{2}}{\frac{5}{2}}$$

$$=\frac{\frac{1}{2}}{\frac{5}{2}}$$

$$=\frac{1}{5}$$

Hence, the correct answer is Option B