A value of θ for which 2 + 3 i sin θ 1 - 2 i sin θ is purely imaginary, is

Let, $z=\frac{\mathrm{2 }+ 3i\sin \theta }{\mathrm{1 }- 2i\sin \theta }$

Multiplying the denominator by its conjugate, we get,

$z=\frac{\left(2+3 i\sin \theta \right) \left(1+2 i\sin \theta \right)}{1+4 {\sin }^2 \theta }$

$=\frac{\left(2-\mathrm{6 }{\sin }^2 \theta \right)+i\left(7\sin \theta \right)}{1+\mathrm{4 }{\sin }^2 \theta }$

To make it purely imaginary. Its real part should be $0$.

Hence, $2=\mathrm{6 }{\sin }^2\theta$

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