If $z=\cos \theta + i \sin \theta$, express $\displaystyle \frac {1}{1-z \cos \theta}$ in the form $a+i\cdot b$. In order to get rid of the $z$, should I substitute $z=\cos\theta + i\sin \theta$ into the complex number, or what conjugate should I multiply the complex number by?
I have tried substituting $z=\cos\theta + i\sin \theta$ into the complex number, but only got this far:
$\displaystyle \frac {1}{1-z \cos \theta}$
$= \displaystyle\frac{1}{1-(\cos\theta + i\sin\theta)(\cos\theta)}$
$=\displaystyle\frac{1}{1-\cos^2\theta-i\cos\theta\sin\theta}$
As for multiplying the complex number by a conjugate, I have used $\big(\displaystyle\frac {1}{z}-\frac{1}{\cos\theta}\big)$, $\big(\displaystyle z-\frac{1}{z}\big)$ and $(1+z\cos\theta)$ but to no avail.
I have only learnt de Moivre's theorem, and I haven't learnt $\cosθ+i\sinθ=e^{iθ}$, so I would appreciate if this question can be solved in the simplest way possible. But other methods are welcome.

\begin{align}
\frac1{1-z\cos\theta}&=\frac1{1-(\cos\theta+i\sin\theta)\cos\theta}\\
&=\frac1{1-\cos^2\theta-i\sin\theta\cos\theta}\\
&=\frac1{\sin^2\theta-i\sin\theta\cos\theta}\color{blue}{\cdot\frac{\sin^2\theta+i\sin\theta\cos\theta}{\sin^2\theta+i\sin\theta\cos\theta}}\\
&=\frac{\sin^2\theta+i\sin\theta\cos\theta}{\sin^4\theta+\sin^2\theta\cos^2\theta}\\
&=\frac{\sin^2\theta+i\sin\theta\cos\theta}{\sin^2\theta(\sin^2\theta+\cos^2\theta)}\\
&=\frac{\sin^2\theta}{\sin^2\theta}+\frac{i\sin\theta\cos\theta}{\sin^2\theta}\\
&=1+i\cot\theta
\end{align}
