Let A={θ ∈(0,2 π): (1+2 i sin θ)/(1-i sin θ). is purely imaginary }. Then the sum of the elements in A is

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