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  • The real value of α for which the expression ((1-isin α))/(1+2i sin⁡α ) is purely real is:

The real value of α for which the expression \frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha} is purely real is:

A.\left ( n+1 \right )\frac{\pi}{2}

B.\left ( 2n+1 \right )\frac{\pi}{2}

C.n\pi

D. None of these

Answers (1)

The answer is the option (c).

Let  z=\frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha}

\begin{aligned} &\begin{array}{c} =\frac{(1-i\sin \alpha)(1-2 i \sin \alpha)}{(1+2 i \sin \alpha)(1-2 i \sin \alpha)} \\\\ =\frac{1-2 i \sin \alpha-i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{(1)^{2}-(2 i \sin \alpha)^{2}} \\\\ =\frac{1-3 i \sin \alpha-2 \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha}=\frac{\left(1-2 \sin ^{2} \alpha\right)-3 i \sin \alpha}{1+4 \sin ^{2} \alpha} \\\\ \frac{-3 i \sin \alpha}{1+4 \sin ^{2} \alpha}=0 \\\\ \sin \alpha=0 \quad \operatorname{} \alpha=n \pi \end{array}\\ &\text { Hence, c is correct. } \end{aligned}

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