Question

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

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

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

(c) nπ

(d) none of these where n ∈ N.

Answer 1

Given $\frac{1-i\sin \alpha }{1+2i\sin \alpha }$ is purely real
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{1-i\sin \alpha }{1+2i\sin \alpha }\times \left(\frac{1-2i\sin \alpha }{1-2i\sin \alpha }\right) \\ =\frac{1-i\sin \alpha -2i\sin \alpha +2i^2{\sin }^2\alpha }{1-4i^2{\sin }^2\alpha } \\ =\frac{1-3i\sin \alpha -2{\sin }^2\alpha }{1+4{\sin }^2\alpha } \\ =\frac{1-2{\sin }^2\alpha }{1+4{\sin }^2\alpha }+i\left(\frac{-3\sin \alpha }{1+4{\sin }^2\alpha }\right) \end{gathered}$$

Which is given to purely real

$$\begin{gathered} \Rightarrow \frac{-3\sin \alpha }{1+4{\sin }^2\alpha }=0 \\ \Rightarrow -3\sin \alpha =0 \\ \mathrm{i}.\mathrm{e}\sin \alpha =0 \\ \mathrm{i}.\mathrm{e}\alpha =n\pi \end{gathered}$$
Hence, the correct answer is option C.