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

$$4 + 5 \tan \theta = \sec \theta$$.

Square both sides: $$(4 + 5 \tan \theta)^2 = \sec^2 \theta = 1 + \tan^2 \theta$$.

$$16 + 40 \tan \theta + 25 \tan^2 \theta = 1 + \tan^2 \theta$$.

$$24 \tan^2 \theta + 40 \tan \theta + 15 = 0$$.

Using the quadratic formula for $$\tan \alpha$$:

$$\tan \alpha = \frac{-40 \pm \sqrt{1600 - 4(24)(15)}}{2(24)} = \frac{-40 \pm \sqrt{1600 - 1440}}{48} = \frac{-40 \pm \sqrt{160}}{48} = \frac{-40 \pm 4\sqrt{10}}{48} = \frac{-10 \pm \sqrt{10}}{12}$$.

the value is $$\frac{\sqrt{10}-10}{12}$$