Two tangents are drawn from a point $$(-2, -1)$$ to the curve, $$y^2 = 4x$$. If $$\alpha$$ is the angle between them, then $$|\tan\alpha|$$ is equal to:

We are given a point $$(-2, -1)$$ and the parabola $$y^2 = 4x$$. We need to find $$|\tan \alpha|$$, where $$\alpha$$ is the angle between the two tangents drawn from this point to the parabola.

First, recall that for a parabola $$y^2 = 4ax$$, the equation of a tangent with slope $$m$$ is $$y = mx + \frac{a}{m}$$. Comparing $$y^2 = 4x$$ with $$y^2 = 4ax$$, we get $$4a = 4$$, so $$a = 1$$. Therefore, the equation of a tangent to the parabola $$y^2 = 4x$$ with slope $$m$$ is:

$$y = mx + \frac{1}{m}$$

Since the tangent passes through the point $$(-2, -1)$$, we substitute $$x = -2$$ and $$y = -1$$ into the tangent equation:

$$-1 = m \cdot (-2) + \frac{1}{m}$$

Simplify the equation:

$$-1 = -2m + \frac{1}{m}$$

To eliminate the denominator, multiply both sides by $$m$$:

$$-1 \cdot m = \left(-2m + \frac{1}{m}\right) \cdot m$$

This gives:

$$-m = -2m^2 + 1$$

Bring all terms to one side to form a quadratic equation:

$$2m^2 - m - 1 = 0$$

Solve for $$m$$ using the quadratic formula. Here, $$a = 2$$, $$b = -1$$, and $$c = -1$$. The discriminant $$D$$ is:

$$D = b^2 - 4ac = (-1)^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9$$

So, the solutions are:

$$m = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-1) \pm \sqrt{9}}{2 \cdot 2} = \frac{1 \pm 3}{4}$$

Thus, the two slopes are:

$$m_1 = \frac{1 + 3}{4} = \frac{4}{4} = 1$$

$$m_2 = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2}$$

The angle $$\alpha$$ between two lines with slopes $$m_1$$ and $$m_2$$ is given by:

$$\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$

Substitute $$m_1 = 1$$ and $$m_2 = -\frac{1}{2}$$:

$$m_1 - m_2 = 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2}$$

$$1 + m_1 m_2 = 1 + (1) \cdot \left(-\frac{1}{2}\right) = 1 - \frac{1}{2} = \frac{1}{2}$$

Therefore,

$$\tan \alpha = \left| \frac{\frac{3}{2}}{\frac{1}{2}} \right| = \left| \frac{3}{2} \times \frac{2}{1} \right| = |3| = 3$$

Hence, $$|\tan \alpha| = 3$$.

Comparing with the options, we see that 3 corresponds to option D.

Hence, the correct answer is Option D.