Let $$A = \left\{\theta \in (0, 2\pi) : \frac{1 + 2i\sin\theta}{1 - i\sin\theta} \text{ is purely imaginary}\right\}$$. Then the sum of the elements in $$A$$ is

We need to find the set $$A = \left\{\theta \in (0, 2\pi) : \frac{1 + 2i\sin\theta}{1 - i\sin\theta} \text{ is purely imaginary}\right\}$$ and then compute the sum of its elements.

To simplify, rationalize the expression.

Multiply the numerator and denominator by the conjugate of the denominator:

$$\frac{1 + 2i\sin\theta}{1 - i\sin\theta} \times \frac{1 + i\sin\theta}{1 + i\sin\theta} = \frac{(1 + 2i\sin\theta)(1 + i\sin\theta)}{(1)^2 + (\sin\theta)^2}$$

Expanding the numerator.

$$= \frac{1 + i\sin\theta + 2i\sin\theta + 2i^2\sin^2\theta}{1 + \sin^2\theta}$$

$$= \frac{(1 - 2\sin^2\theta) + i(3\sin\theta)}{1 + \sin^2\theta}$$

Now apply the condition for purely imaginary.

For the expression to be purely imaginary, the real part must be zero and the imaginary part must be non-zero:

Real part = 0:

$$\frac{1 - 2\sin^2\theta}{1 + \sin^2\theta} = 0$$

$$1 - 2\sin^2\theta = 0$$

$$\sin^2\theta = \frac{1}{2}$$

$$\sin\theta = \pm\frac{1}{\sqrt{2}}$$

Imaginary part ≠ 0:

$$\frac{3\sin\theta}{1 + \sin^2\theta} \neq 0$$

This requires $$\sin\theta \neq 0$$, which is satisfied for all solutions above.

Now, find all values of $$\theta$$ in $$(0, 2\pi)$$.

From $$\sin\theta = \frac{1}{\sqrt{2}}$$: $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}$$

From $$\sin\theta = -\frac{1}{\sqrt{2}}$$: $$\theta = \frac{5\pi}{4}, \frac{7\pi}{4}$$

So $$A = \left\{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\right\}$$

Now, compute the sum.

$$\frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi$$

The sum of the elements in $$A$$ is $$4\pi$$, which corresponds to Option A.