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TS EAMCET · Maths · Complex Number

The set of all values of \(\theta\) such that \(\frac{1-i \cos \theta}{1+2 i \sin \theta}\) is purely imaginary is

  1. A \(\left\{n \pi+(-1)^n \frac{\pi}{4}, n \in z\right\}\)
  2. B \(\left\{\frac{n \pi}{2}+(-1)^n \frac{\pi}{4}, n \in z\right\}\)
  3. C \(\left\{n \pi+(-1)^n \frac{\pi}{2}, n \in z\right\}\)
  4. D \(\left\{2 n \pi \pm \frac{\pi}{4}, n \in z\right\}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left\{\frac{n \pi}{2}+(-1)^n \frac{\pi}{4}, n \in z\right\}\)

Step-by-step Solution

Detailed explanation

\(\frac{1-i \cos \theta}{1+2 i \sin \theta} = \frac{(1-i \cos \theta)(1-2 i \sin \theta)}{1+4 \sin^2 \theta} = \frac{1-2 i \sin \theta-i \cos \theta-2 \sin \theta \cos \theta}{1+4 \sin^2 \theta}\) \(\text{Real part} = \frac{1-2 \sin \theta \cos \theta}{1+4 \sin^2 \theta}\)…
From TS EAMCET
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