The real value of α for which the expression


The real value of $\alpha$ for which the expression $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real, is

(a) $(n+1) \frac{\pi}{2}$

(b) $(2 n+1) \frac{\pi}{2}$


(d) none of these where n ∈ N.


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 $\mathrm{C}$.


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