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