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 :

$$ \begin{aligned} & \text { Here, } z=\frac{1+2 i \sin \theta}{1-i \sin \theta} \times \frac{1+i \sin \theta}{1+i \sin \theta} \\\\ & \frac{1+i \sin \theta+2 i \sin \theta-2 \sin ^2 \theta}{1-i^2 \sin ^2 \theta} \\\\ & =\frac{\left(1-2 \sin ^2 \theta\right)+i(3 \sin \theta)}{1+\sin ^2 \theta} \end{aligned} $$ <br/><br/>$\because z$ is purely imaginary, so $\operatorname{Re} z=0$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{1-2 \sin ^2 \theta}{1+\sin ^2 \theta}=0 \\\\ & \Rightarrow 2 \sin ^2 \theta=1 \Rightarrow \sin ^2 \theta=\frac{1}{2} \\\\ & \Rightarrow \sin \theta= \pm \frac{1}{\sqrt{2}} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \therefore A=\left[\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right] \because \theta \in(0,2 \pi)\\\\ & \therefore \text { Sum }=\frac{\pi+3 \pi+5 \pi+7 \pi}{4}=\frac{16 \pi}{4}=4 \pi \end{aligned} $$