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The real value of $\alpha$ for which the expression $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real is
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Verified Answer
The correct answer is:
$n \pi$
$n \pi$
Let, $z=\frac{1-i \sin \alpha}{1+2 i \sin \alpha}=\frac{(1-i \sin \alpha)(1-2 i \sin \alpha)}{(1+2 i \sin \alpha)(1-2 i \sin \alpha)}$ $=\frac{1-i \sin \alpha-2 i \sin \alpha+2 i^2 \sin ^2 \alpha}{1-4 i^2 \sin ^2 \alpha}$
$$
=\frac{1-3 i \sin \alpha-2 \sin ^2 \alpha}{1+4 \sin ^2 \alpha}=\frac{1-2 \sin ^2 \alpha}{1+4 \sin ^2 \alpha}-\frac{3 i \sin \alpha}{1+4 \sin ^2 \alpha}
$$
As, $\mathrm{z}$ is a purely real.
So, $\frac{-3 \sin \alpha}{1+4 \sin ^2 \alpha}=0 \Rightarrow \sin \alpha=0 \Rightarrow \alpha=n \pi$
$$
=\frac{1-3 i \sin \alpha-2 \sin ^2 \alpha}{1+4 \sin ^2 \alpha}=\frac{1-2 \sin ^2 \alpha}{1+4 \sin ^2 \alpha}-\frac{3 i \sin \alpha}{1+4 \sin ^2 \alpha}
$$
As, $\mathrm{z}$ is a purely real.
So, $\frac{-3 \sin \alpha}{1+4 \sin ^2 \alpha}=0 \Rightarrow \sin \alpha=0 \Rightarrow \alpha=n \pi$
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