The real value of α for which the expression

Given $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real

i. e $\frac{1-i \sin \alpha}{1+2 i \sin \alpha} \times\left(\frac{1-2 i \sin \alpha}{1-2 i \sin \alpha}\right)$

$=\frac{1-i \sin \alpha-2 i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha}$

$=\frac{1-3 i \sin \alpha-2 \sin ^{2} \alpha}{1+4 \sin ^{2} \alpha}$

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

Which is given to purely real

$\Rightarrow \frac{-3 \sin \alpha}{1+4 \sin ^{2} \alpha}=0$

$\Rightarrow-3 \sin \alpha=0$

i. e $\sin \alpha=0$

i. e $\alpha=n \pi$

Hence, the correct answer is option $\mathrm{C}$.