Prove that $\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$ 
Prove that $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$$

I tried to rationalize the denominator but I always end up with a large fraction that won't cancel out. Is there something I'm missing?
Thanks in advance
 A: This is just a matter of observing that\begin{align}(1+\sin\theta-i\cos\theta)(\sin\theta+i\cos\theta)&=\sin\theta+\sin^2\theta+\cos^2\theta+i(\cos\theta+\sin\theta\cos\theta-\cos\theta\sin\theta)\\&=1+\sin\theta+i\cos\theta.\end{align}
A: $(e^{j*0} + e^{-j*\theta})*e^{j*\theta} = 1 + e^{j*\theta}$
Much simpler would be to convert to polar coordinates.
A: $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\frac{2\cos^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)+2i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)-2i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=$$
$$=\frac{\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)+i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)-i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\left(\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)+i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)^2=\sin\theta+i\cos\theta.$$
I used $$1+\cos\alpha=2\cos^2\frac{\alpha}{2};$$
$$\sin\alpha=2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}$$ and
$$\sin\alpha=\cos\left(\frac{\pi}{2}-\alpha\right).$$
A: Let $z=\sin\theta+i\cos\theta.\;$ Then $\bar{z}=\sin\theta-i\cos\theta.\;$ Now $z\bar{z}=\sin^2\theta+\cos^2\theta=1.$ Thus, $
(1+\sin\theta+i\cos\theta)/(1+\sin\theta-i\cos\theta)=( 1+z)/(z\bar{z}+\bar{z})=(1+z)/(\bar{z}(z+1))=z.$
A: We know that, $\sin^2\theta+\cos^2\theta = 1$ and 
$a^2-b^2=(a-b)(a+b$
then 
\begin{split} \frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} &= &\frac{\color{red}{\sin^2\theta+\cos^2\theta} +\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} \qquad\quad\\\\&=& \frac{\color{red}{\sin^2\theta+(-i\cos\theta)(i\cos\theta)} +\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} \\
\\ &=&\frac{\color{red}{[\sin^2\theta- (i\cos\theta)^2]}+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}  \\\qquad~~~\qquad\\&=&\frac{\color{red}{(\sin\theta +i\cos\theta)(\sin\theta- i\cos\theta)}+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} \\\\&=&(\sin\theta +i\cos\theta)\frac{1+\sin\theta- i\cos\theta}{1+\sin\theta-i\cos\theta} \\&=&(\sin\theta +i\cos\theta) \end{split}
A: Let $\phi=\pi/2-\theta$. Then
$$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}
=\frac{1+\cos\phi+i\sin\phi}{1+\cos\phi-i\sin\phi}=\frac{1+e^{i\phi}}
{1+e^{-i\phi}}$$
and
$$\sin\theta+i\cos\theta=\cos\phi+i\sin\phi=e^{i\phi}$$
etc.
A: write $$\frac{(1+\sin(\theta)+i\cos(\theta))^2}{(1+\sin(\theta)-i\cos(\theta))(1+\sin(\theta)+i\sin(\theta))}$$
simplifying the denominator we obtain
$$1+\sin^2(\theta)+2\sin(\theta)-\cos^2(\theta)+2i(1+\sin(\theta))\cos(\theta)$$
and the denominator $$1+\sin^2(\theta)+2\sin(\theta)-\cos^2(\theta)=2(1+\sin(\theta)-\cos^2(\theta))$$
A: Consider that
$$
\sin\theta+i\cos\theta=ie^{-i\theta}\\
\sin\theta-i\cos\theta=-ie^{i\theta}
$$
Then, does
$$\frac{1+ie^{-i\theta}}{1-ie^{i\theta}}=ie^{-i\theta}\\
\text{or}\\
1+ie^{-i\theta}=ie^{-i\theta}(1-ie^{i\theta})=ie^{-i\theta}+1
$$
YES, as was to be determined.
