Question

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

• A. 1/(3 + 5cos θ)
• B. 1/(5 - 3cos θ)
• C. 1/(3 - 5cos θ)
• D. 1/(5 + 3cos θ)

Answer 1

Answer: (D)

Given:

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

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

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

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

Real part $=\frac{(1-\cos \theta )}{(1-\cos \theta {)}^2+4{\sin }^2\theta }$

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

$=\frac{(1-\cos \theta )}{5-3{\cos }^2\theta -2\cos \theta }$

$=\frac{(1-\cos \theta )}{(1-\cos \theta )(5+3\cos \theta )}$

$=\frac{1}{(5+3\cos \theta )}$

Therefore, The real part of $(1-\cos \theta +2i\sin \theta {)}^{-1}$ is $\frac{1}{(5+3\cos \theta )}$.

Hence, Option D is correct.