Answer:
Option D
Explanation:
Let $z= \frac{2+3i\sin\theta}{1-2i\sin\theta}$
is purely
imaaginary Then , we have Re(z)=0
NOw conside $z= \frac{2+3i\sin\theta}{1-2i\sin\theta}$
$z= \frac{(2+3i\sin\theta)(1+2i\sin\theta)}{(1-2i\sin\theta)(1+2i\sin\theta)}$
$z= \frac{(2+4i\sin\theta+3i\sin\theta+6i^{2}\sin^{2}\theta)}{(1^{2}-(2i\sin\theta)^{2}}$
$z= \frac{(2+7i\sin\theta-6\sin^{2}\theta)}{(1+4\sin^{2}\theta)}$
$z= \frac{(2-6\sin^{2}\theta)}{(1+4\sin^{2}\theta)}+i\frac{7\sin\theta}{(1+4\sin^{2}\theta)}$
Re(z)=0
$\frac{(2-6\sin^{2}\theta)}{(1+4\sin^{2}\theta)}=0$
$2=6\sin^{2}\theta$
$\Rightarrow sin^{2}\theta=\frac{1}{\sqrt{3}}$
$\Rightarrow sin^{2}\theta=\pm\frac{1}{\sqrt{3}}$
$\Rightarrow \theta= \sin^{-}(\pm \frac{1}{\sqrt{3}})=\pm sin^{-}(\frac{1}{\sqrt{3}})$
