If $\frac{2 + 3\mathrm{isin\theta }}{1 - 2\mathrm{i} \mathrm{sin\theta }}$ is purely imaginary, then ${\cos }^2\mathrm{\theta } =$

$$\begin{gathered} \frac{2+3\mathrm{i} \sin \mathrm{\theta }}{1-2\mathrm{i} \sin \mathrm{\theta } }=\frac{(2+3\mathrm{i} \mathrm{sin\theta }) (1+2\mathrm{i} \mathrm{sin\theta })}{(1-2 \mathrm{i} \mathrm{sin\theta }) (1+2\mathrm{i} \mathrm{sin\theta })} \\ =\frac{(2-6{\sin }^2\mathrm{\theta })+\mathrm{i}\left(7\mathrm{sin\theta }\right)}{1+4{\sin }^2 \mathrm{\theta }} \\ \mathrm{is} \mathrm{purely} \mathrm{imaginary} \\ \Rightarrow \mathrm{Real} \mathrm{part} \mathrm{of} \frac{2+3\mathrm{isin\theta }}{1-2\mathrm{isin\theta }}=0 \\ \Rightarrow 2-6{\sin }^2\mathrm{\theta }=0 \Rightarrow s{\mathrm{in}}^2\mathrm{\theta }=\frac{1}{3} \\ \boxed{\Rightarrow {\cos }^2\mathrm{\theta }=\frac{2}{3}} \end{gathered}$$