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 $\frac{1}{2}$

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

$=\frac{1}{2{\sin }^2\frac{\theta }{2}+i\cdot 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}=\frac{1}{2\sin \frac{\theta }{2}}\cdot \frac{1}{\sin \frac{\theta }{2}+i\cos \frac{\theta }{2}}$

$=\frac{1}{2\sin \frac{\theta }{2}}\cdot \frac{1}{\sin \frac{\theta }{2}+i\cos \frac{\theta }{2}}\cdot \frac{\sin \frac{\theta }{2}-i\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}-i\cos \frac{\theta }{2}}$, rationalize denominator

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

$=\frac{1}{2}-\frac{i}{2}\cot \frac{\theta }{2}$

The real part is $\frac{1}{2}$