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COMEDK · Maths · 3. Complex Number

The modulus of \([1-\cos \theta+i \sin \theta]^{-1}\) is

  1. A \(\frac{1}{2} \operatorname{cosec} \frac{\theta}{2}\)
  2. B \(\operatorname{cosec} \frac{\theta}{2}\)
  3. C \(\frac{1}{2}\left|\operatorname{cosec} \frac{\theta}{2}\right|\)
  4. D \(\left|\operatorname{cosec} \frac{\theta}{2}\right|\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{2}\left|\operatorname{cosec} \frac{\theta}{2}\right|\)

Step-by-step Solution

Detailed explanation

\begin{aligned} &\left|\frac{1}{1-\cos \theta+i \sin \theta}\right|=\frac{1}{|1-\cos \theta+i \sin \theta|} \\ &=\frac{1}{\sqrt{(1-\cos \theta)^{2}+\sin ^{2} \theta}}=\frac{1}{\sqrt{2-2 \cos \theta}} \\ &=\frac{1}{2|\sin (\theta / 2)|}=\frac{1}{2}\left|\operatorname{cosec}…