Question

Express $\frac{1}{1-cos\theta +2isin\theta }$ in the form of a + ib.

Answer 1

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

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

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

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

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