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

We have, $(1- \cos \,\theta + i \,\sin \,\theta)^{-1} $ $=\frac{1}{1-\cos\, \theta+i\, \sin\, \theta} $ $=\frac{1-\cos\, \theta-i \,\sin \,\theta}{(1-\cos \,\theta)^{2}+\sin ^{2} \theta}$ $=\frac{1-\cos\, \theta-i \,\sin\, \theta}{1-2 \,\cos \,\theta+\cos ^{2} \theta+\sin ^{2} \theta}$ $=\frac{1-\cos\, \theta-i \\,sin\, \theta}{2-2 \,\cos \,\theta}$ $=\frac{1-\cos\, \theta}{2(1-\cos \,\theta)}-\frac{i\, \sin\, \theta}{2(1-\cos \,\theta)} $ $=\frac{1}{2}-\frac{i \,\sin\, \theta}{2(1-\cos\, \theta)}$ Hence, the real part is $\frac{1}{2}$.