Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{1+\cos \theta }$
• C. $\tan \frac{\theta }{2}$
• D. $\cot \frac{\theta }{2}$

Answer 1

The correct option is A $\frac{1}{2}$

$\frac{1}{1-\cos \theta +i\sin \theta }$
$=\frac{1-\cos \theta -i\sin \theta }{(1-\cos \theta {)}^2-(i\sin \theta {)}^2}$
$=\frac{1-\cos \theta -i\sin \theta }{1-2\cos \theta +{\cos }^2\theta +{\sin }^2\theta }$
$=\frac{1}{2}-\frac{i}{2}\frac{\sin \theta }{1-\cos \theta }$