If $$\alpha$$, $$-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$$ is the solution of $$4\cos\theta + 5\sin\theta = 1$$, then the value of $$\tan\alpha$$ is

A

$$\frac{10 - \sqrt{10}}{6}$$

B

$$\frac{10 - \sqrt{10}}{12}$$

C

$$\frac{\sqrt{10} - 10}{12}$$

D

$$\frac{\sqrt{10} - 10}{6}$$