Question

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)
(A) \((2 n+1) \frac{\pi}{2}, n \in I\)
(B) \(\frac{3 n \pi}{2}, n \in I\)
(C) \(n \pi, n \in 1\)
(D) \(2 n \pi, n \in I\)

Answer 1

Ans: (A)
$$
\text { Hint } \begin{aligned}
\because \frac{1-i \cos \theta}{1+2 i \cos \theta} \text { is real } & \Rightarrow \frac{1-i \cos \theta}{1+2 i \cos \theta}=\frac{1+i \cos \theta}{1-2 i \cos \theta} \\
& \Rightarrow 1-3 i \cos \theta-2 \cos ^{2} \theta=1+3 i \cos \theta-2 \cos ^{2} \theta \quad \Rightarrow \cos \theta=0 \\
& \Rightarrow \theta=n \pi+\pi / 2(n \in I)
\end{aligned}
$$

Answer 2

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Answer 3

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