Let A = { 0 ∈ ( -π/2 , π) : 3 + 2isinθ/1 - 2isinθ is purely imaginary}

Given $\mathrm{z}=\frac{3+2 \mathrm{i} \sin \theta}{1-2 \mathrm{i} \sin \theta}$ is purely img

so real part becomes zero.

$\mathrm{z}=\left(\frac{3+2 \mathrm{i} \sin \theta}{1-2 \mathrm{i} \sin \theta}\right) \times\left(\frac{1+2 \mathrm{i} \sin \theta}{1+2 \mathrm{i} \sin \theta}\right)$

$z=\frac{\left(3-4 \sin ^{2} \theta\right)+i(8 \sin \theta)}{i+4 \sin ^{2} \theta}$

Now $\operatorname{Re}(\mathrm{z})=0$

$\frac{3-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}=0$

$\sin ^{2} \theta=\frac{3}{4}$

$\sin \theta=\pm \frac{\sqrt{3}}{2} \Rightarrow \theta=--\frac{\pi}{3}, \frac{\pi}{3}, \frac{2 \pi}{3}$

$\because \quad \theta \in\left(-\frac{\pi}{2}, \pi\right)$

then sum of the elements in $\mathrm{A}$ is

$-\frac{\pi}{3}+\frac{\pi}{3}+\frac{2 \pi}{3}=\frac{2 \pi}{3}$