If $\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is a real number and $0<\theta<2 \pi$, then $\theta=$
(a) $\pi$
(b) $\frac{\pi}{2}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{6}$
(a) $\pi$
Given:
$\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is a real number
On rationalising, we get,
$\frac{3+2 i \sin \theta}{1-2 i \sin \theta} \times \frac{1+2 i \sin \theta}{1+2 i \sin \theta}$
$=\frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1)^{2}-(2 i \sin \theta)^{2}}$
$=\frac{3+2 i \sin \theta+6 i \sin \theta+4 i^{2} \sin ^{2} \theta}{1+4 \sin ^{2} \theta}$
$=\frac{3-4 \sin ^{2} \theta+8 i \sin \theta}{1+4 \sin ^{2} \theta} \quad\left[\because i^{2}=-1\right]$
$=\frac{3-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}+i \frac{8 \sin \theta}{1+4 \sin ^{2} \theta}$
For the above term to be real, the imaginary part has to be zero.
$\therefore \frac{8 \sin \theta}{1+4 \sin ^{2} \theta}=0$
$\Rightarrow 8 \sin \theta=0$
For this to be zero,
$\sin \theta=0$
$\Rightarrow \theta=0, \pi, 2 \pi, 3 \pi \ldots$
But $0<\theta<2 \pi$
Hence, $\theta=\pi$