The real value of ' \(\alpha\) ' for which \(\frac{1-i \sin \alpha}{1+2 i \sin \alpha}\) is purely real is

Given, \(\frac{1-i \sin \alpha}{1+2^j \sin \alpha}\) is purely real
i.e. \(\frac{1-i \sin \alpha}{1+2 i \sin \alpha} \times \frac{(1-2 i \sin \alpha)}{(1-2 i \sin \alpha)}\)
\(=\frac{1-i \sin \alpha-2 i \sin \alpha+2 i^2 \sin ^2 \alpha}{1-4 i^2 \sin ^2 \alpha}\)
\(=\frac{1-3 i \sin \alpha-2 \sin ^2 \alpha}{1+4 \sin ^2 \alpha}\)
\(=\frac{1-2 \sin ^2 \alpha}{1+4 \sin ^2 \alpha}+i\left(\frac{-3 \sin \alpha}{1+4 \sin ^2 \alpha}\right)\)
which is given to purely real
\(\Rightarrow \quad \frac{-3 \sin \alpha}{1+4 \sin ^2 \alpha}=0\)
\(\Rightarrow \quad-3 \sin \alpha=0 \Rightarrow \sin \alpha=0\)
Hence, \(\alpha=n \pi, n \in N\)