Question

The real part of ${\left(1-\cos \theta +2i\sin \theta \right)}^{-1}$ is

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

Answer 1

The correct option is D $\frac{1}{5+3\cos \theta }$

$\frac{1}{1-\cos \theta +2i\sin \theta }=\frac{(1-\sin \theta )-2i\sin \theta }{{(1-\cos \theta )}^2+4{\sin }^2\theta }=P$ (Let)

$Re\left(P\right)=\frac{1-\cos \theta }{{(1-\cos \theta )}^2+4{\sin }^2\theta }=\frac{1-\cos \theta }{1+{\cos }^2\theta -2\cos \theta +4{\sin }^2\theta }$

$Re\left(P\right)=\frac{1-\cos \theta }{1+3{\sin }^2\theta -2\cos \theta +1}=\frac{1-\cos \theta }{2+3-3{\cos }^2\theta -2\cos \theta }$

$Re\left(P\right)=\frac{1-\cos \theta }{5-3{\cos }^2\theta -2\cos \theta }=\frac{1-\cos \theta }{-3{\cos }^2\theta -2\cos \theta +5}$

Using Factorization in denominator

$Re\left(Z\right)=\frac{1-\cos \theta }{(1-\cos \theta )(3\cos \theta +5)}=\frac{1}{(3\cos \theta +5)}$