If $$2 \cos^2 \theta - 5 \cos \theta + 2 = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$(\sec \theta + \tan \theta)$$ is:

$$2\cos^2\theta-5\cos\theta+2=0$$

$$2\cos^2\theta-4\cos\theta-\cos\theta+2=0$$

$$2\cos\theta\left(\cos\theta-2\right)-1\left(\cos\theta-2\right)=0$$

$$\left(\cos\theta-2\right)\left(2\cos\theta-1\right)=0$$

$$\cos\theta=2$$  or $$\cos\theta=\frac{1}{2}$$

$$\cos\theta=2$$ is not possible.

So, $$\cos\theta=\frac{1}{2}$$

$$\sec\theta=2$$

$$\tan\theta=\sqrt{\sec^2\theta-1}=\sqrt{4-1}=\sqrt{3}$$

$$\therefore$$  $$\left(\sec\theta+\tan\theta\right)=2+\sqrt{3}$$

Hence, the correct answer is Option A