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If $\theta \in \mathbb{R}$ and $\frac{1-i \cos \theta}{1+2 i \cos \theta}$ is real number, then $\theta$ will be (when $I:$ Set of integers)
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The correct answers are:
$(2 n+1) \frac{\pi}{2}, n \in l$
Let $Z=\frac{1-i \cos \theta}{1+2 i \cos \theta}$
$\therefore \quad \bar{z}=\frac{1+i \cos \theta}{1-2 i \cos \theta}$
Since, $z$ is a real number, then $z-\bar{z}=0$ $\Rightarrow \frac{1-i \cos \theta}{1+2 i \cos \theta}=\frac{1+i \cos \theta}{1-2 i \cos \theta}$
$\Rightarrow(1-i \cos \theta)(1-2 i \cos \theta)=(1+i \cos \theta)(1+2 i \cos \theta)$
$\Rightarrow 1-2 i \cos \theta-i \cos \theta-2 \cos ^{2} \theta$ $=1+2 i \cos \theta+i \cos \theta-2 \cos ^{2} \theta$
$\Rightarrow \quad 6 i \cos \theta=0$
$\Rightarrow \quad \cos \theta=0$
$\Rightarrow \quad \theta=(2 n+1) \frac{\pi}{2}, n \in I$
$\therefore \quad \bar{z}=\frac{1+i \cos \theta}{1-2 i \cos \theta}$
Since, $z$ is a real number, then $z-\bar{z}=0$ $\Rightarrow \frac{1-i \cos \theta}{1+2 i \cos \theta}=\frac{1+i \cos \theta}{1-2 i \cos \theta}$
$\Rightarrow(1-i \cos \theta)(1-2 i \cos \theta)=(1+i \cos \theta)(1+2 i \cos \theta)$
$\Rightarrow 1-2 i \cos \theta-i \cos \theta-2 \cos ^{2} \theta$ $=1+2 i \cos \theta+i \cos \theta-2 \cos ^{2} \theta$
$\Rightarrow \quad 6 i \cos \theta=0$
$\Rightarrow \quad \cos \theta=0$
$\Rightarrow \quad \theta=(2 n+1) \frac{\pi}{2}, n \in I$
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