Question

Real part of $(1-\cos \theta+2 i \sin \theta)^{-1}$ is

• A. $\frac{1}{(3+5 \cos \theta)}$
• B. $\frac{1}{(5+3 \cos \theta)}$
• C. $\frac{1}{(5-3 \cos \theta)}$
• D. $\frac{1}{(3-5 \cos \theta)}$

Answer 1

Answer: (B)

$ z =\frac{1}{1-\cos \theta+2 i \sin \theta} $
$=\frac{1-\cos \theta-2 i \sin \theta}{(1-\cos \theta)^{2}+4 \sin ^{2} \theta} $
Real part $=\frac{1-\cos \theta}{(1-\cos \theta)^{2}+4\left(1-\cos ^{2} \theta\right)}$
$=\frac{1}{1-\cos \theta+4(1+\cos \theta)}$
$=\frac{1}{5+3 \cos \theta}$