1/(1-cos theta +2i sin theta) please convert it in x+iy form

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$$\begin{gathered} To Express : \frac{1}{1-\cos \theta +2i\sin \theta } as x+iy \\ solution : \frac{1}{(1-\cos \theta )+2i\sin \theta }\times \frac{(1-\cos \theta )-2i\sin \theta }{(1-\cos \theta )-2i\sin \theta } \\ =\frac{(1-\cos \theta )-2i\sin \theta }{(1-\cos \theta {)}^2-(2i\sin \theta {)}^2} \\ =\frac{(1-\cos \theta )-2i\sin \theta }{(1^2+{\cos }^2\theta -2\cos \theta )-\left(2^2{i}^2{\sin }^2\theta \right)} \\ =\frac{(1-\cos \theta )-2i\sin \theta }{1+{\cos }^2\theta -2\cos \theta +4{\sin }^2\theta } \\ =\frac{(1-\cos \theta )-2i\sin \theta }{1+{\cos }^2\theta +{\sin }^2\theta -2\cos \theta +3{\sin }^2\theta } \\ =\frac{(1-\cos \theta )-2i\sin \theta }{2-2\cos \theta +3{\sin }^2\theta } \\ =\frac{1-\cos \theta }{2-2\cos \theta +3{\sin }^2\theta }+i\frac{-2\sin \theta }{2-2\cos \theta +3{\sin }^2\theta } \\ Note : Here, it is not give to use "a \cos \theta +i a \sin \theta \Rightarrow \sqrt{x^2+y^2}=a and \theta ={\tan }^{-1}\frac{y}{x}". We only have to express as real terms seperate and imaginary terms seperate that\text{'}s it. \\ So, comapring above expression with x+iy \\ \Rightarrow x=\frac{1-\cos \theta }{2-2\cos \theta +3{\sin }^2\theta } and y=\frac{-2\sin \theta }{2-2\cos \theta +3{\sin }^2\theta } \\ Hence, \frac{1}{1-\cos \theta +2i\sin \theta } can be expressed as \frac{1-\cos \theta }{2-2\cos \theta +3{\sin }^2\theta }+i\frac{-2\sin \theta }{2-2\cos \theta +3{\sin }^2\theta } as x+iy form. \end{gathered}$$

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