The real value of θ for which the expression

Given $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is a real number

i. e $\frac{1+i \cos \theta}{1-2 i \cos \theta} \times \frac{1+2 i \cos \theta}{1+2 i \cos \theta}$

$=\frac{(1+i+\cos \theta)(1+2 i \cos \theta)}{1-4 i^{2} \cos ^{2} \theta}$

$=\frac{1+2 i \cos \theta+i \cos \theta+2 i^{2} \cos ^{2} \theta}{1+4 \cos ^{2} \theta}$

$=\frac{1-2 \cos ^{2} \theta+3 i \cos \theta}{1+4 \cos ^{2} \theta}$

$=\frac{1-2 \cos ^{2} \theta}{1+4 \cos ^{2} \theta}+i\left(\frac{3 \cos \theta}{1+4 \cos ^{2} \theta}\right)$

Since $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is a real number

$\Rightarrow \frac{3 \cos \theta}{1+4 \cos ^{2} \theta}=0$

i. e $\cos \theta=0$

i. e $\theta=\frac{(2 n \pm 1) \pi}{2}=n \pi \pm \frac{\pi}{2}$

Hence, the correct answer is option C.