Prove: Im$\left(\frac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \frac{2r\sin\theta}{1-2r\cos\theta + r^2}$ Prove the identity: $\operatorname{Im}\left(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \dfrac{2r\sin\theta}{1-2r\cos\theta + r^2}$
My only lead is to assume $r$ and $\theta$ are real.
Then, 
$\operatorname{Im}\left(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \dfrac{r\sin\theta + \sin\theta}{\sin^2+(1-\cos\theta)^2}$
 A: I suspect there is a missing part in the expression, because the following identity holds:
$
\operatorname{Im}\left(\dfrac{1+re^{i\theta}}{1-re^{i\theta}}\right) = \dfrac{2r\sin\theta}{1-2r\cos\theta + r^2}
$
A: RTP: $Im\bigg(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\bigg) = \dfrac{2r\sin\theta}{1-2r\cos\theta + r^2}$=$\frac{2r\sin\theta}{(r-\cos\theta)^2+1-\cos^2\theta}$
$$Im\bigg(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\bigg) = Im\bigg(\frac{(1+r\cos \theta+ri\sin\theta )(1-\cos\theta +i\sin\theta)}{(1-\cos\theta-i\sin\theta)(1-\cos\theta+i\sin\theta)}\bigg)$$
$$=Im\bigg(\frac{1-\cos\theta+i\sin\theta+r\cos\theta-r\cos^2\theta+r\sin\theta\cos\theta \cdot i +r\sin\theta \cdot i-r\sin\theta\cos\theta\cdot i-r\sin^2\theta}{1-\cos\theta +i\sin\theta -\cos\theta+\cos^2\theta-\cos\theta\sin\theta\cdot i -i\sin\theta +\sin\theta\cos\theta\cdot i +\sin^2\theta}\bigg)$$
$$= \frac{\sin\theta+r\sin\theta}{1-2\cos\theta+\cos^2\theta+\sin^2\theta}$$
$$= \frac{(r+1)\sin\theta}{2(1-\cos\theta)}$$
I don't see how to arrive at the identity from here, could someone comment if they see an error/ way to proceed?
