Question

$$\begin{gathered} \frac{1+i \cos \mathrm{θ}}{1-2i \cos \mathrm{θ}}=\frac{\left(1+i \cos \mathrm{θ}\right)\left(1+2i \cos \mathrm{θ}\right)}{\left(1-2i \cos \mathrm{θ}\right)\left(1+2i \cos \mathrm{θ}\right)} \\ =\frac{1-2{\cos }^2\mathrm{θ}+3i\cos \mathrm{θ}}{1+4{\cos }^2\mathrm{θ}} \end{gathered}$$
is a real number only if  $\frac{3\cos \mathrm{θ}}{1+4{\cos }^2\mathrm{θ}}=0$
i.e. if cos$\mathrm{θ}$ = 0
i.e. if  $\mathrm{θ}$= (2n + 1)$\mathrm{π}/2,\mathrm{n}∈ \mathrm{I}$
So (b) is correct alternative.