Q85 / 180 · KCET 2016 · Mathematics
The real part of \( (1-\cos \theta+i \sin \theta)^{-1} \) is
- Correct answerA\( \frac{1}{2} \)
- B(1) \( \frac{1}{1+\cos \theta} \)
- C\( \tan \frac{\theta}{2} \)
- D\( \odot \cot \frac{\theta}{2} \)
Solution
Given that,
\[ \begin{aligned} (1-\cos \theta+i \sin \theta)^{-1}=\frac{1}{1-\cos \theta+i \sin \theta} \\ =\frac{1}{2 \sin ^{2} \frac{\theta}{2}+i\left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right)} \\ =\frac{1}{2 \sin \frac{\theta}{2}} \times \frac{1}{\sin \frac{\theta}{2}+i \cos \frac{\theta}{2}} \\ =\frac{1}{2 \sin \frac{\theta}{2}} \times \frac{\theta}{\sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}} \\ =\frac{\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2}} \\ =\frac{1}{2}-i\left(\frac{1}{2} \cot \frac{\theta}{2}\right) \end{aligned} \]
So, real part of the complex number is \( \frac{1}{2} \)
\[ \begin{aligned} (1-\cos \theta+i \sin \theta)^{-1}=\frac{1}{1-\cos \theta+i \sin \theta} \\ =\frac{1}{2 \sin ^{2} \frac{\theta}{2}+i\left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right)} \\ =\frac{1}{2 \sin \frac{\theta}{2}} \times \frac{1}{\sin \frac{\theta}{2}+i \cos \frac{\theta}{2}} \\ =\frac{1}{2 \sin \frac{\theta}{2}} \times \frac{\theta}{\sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}} \\ =\frac{\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2}} \\ =\frac{1}{2}-i\left(\frac{1}{2} \cot \frac{\theta}{2}\right) \end{aligned} \]
So, real part of the complex number is \( \frac{1}{2} \)
Hence, the correct answer is option: 'A'.