Let $\mathrm{A}=\left\{\theta \in\left(-\frac{\pi}{2}, \pi\right): \frac{3+2 \mathrm{i} \sin \theta}{1-2 \mathrm{i} \sin \theta}\right.$ is purely imaginary $\}$.
Then the sum of the elements in $A$ is:
Correct Option: , 4
Suppose $z=\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$
Since, $z$ is purely imaginary, then $z+\bar{z}=0$
$\Rightarrow \frac{3+2 i \sin \theta}{1-2 i \sin \theta}+\frac{3-2 i \sin \theta}{1+2 i \sin \theta}=0$
$\Rightarrow \frac{(3+2 i \sin \theta)(1+2 i \sin \theta)+(3-2 i \sin \theta)(1-2 i \sin \theta)}{1+4 \sin ^{2} \theta}=0$
$\Rightarrow \sin ^{2} \theta=\frac{3}{4} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2}$
$\Rightarrow \theta=-\frac{\pi}{3}, \frac{\pi}{3}, \frac{2 \pi}{3}$
Now, the sum of elements in $A=-\frac{\pi}{3}+\frac{\pi}{3}+\frac{2 \pi}{3}=\frac{2 \pi}{3}$