Purely imaginary means real part $=0$

$ \frac{2+3 \sin \theta}{1-2 i \sin \theta} \times \frac{1+2 i \sin \theta}{1+2 i \sin \theta} $

$=\frac{2+4 i \sin \theta+3 i \sin \theta-6 \sin ^{2} \theta}{1-(2 i \sin \theta)^{2}}$

$ =\frac{2+7 i \sin \theta-6 \sin ^{2} \theta}{1+4 \sin ^{2} \theta} $

Real part is $=\frac{2-6 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}$

This is equal to zero.

$ \Rightarrow \frac{2-6 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}=0$

So, $2=6 \sin ^{2} \theta$

$\sin ^{2} \theta=\frac{1}{3}$

$ \sin \theta=\pm \frac{1}{\sqrt{3}} $

$ \therefore \theta=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right) $