The real part of \( (1-\cos \theta+i \sin \theta)^{-1} \) is

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 \( \frac{1}{2} \)