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.

Note that $z=e^{i\theta}$, $z=e^{-i\theta}$ and $\cos\theta = \frac{z+\bar z}2$
$$\frac {1}{1-z \cos \theta}=\frac {1}{1-z \cos \theta}\cdot\frac {1-\bar z \cos \theta}{1-\bar z \cos \theta}$$
$$=\frac {1-e^{-i\theta}\cos \theta}{1-(\bar z +z)\cos \theta + \cos^2\theta}$$
$$=\frac {1-(\cos\theta - i\sin\theta)\cos \theta}{1-2\cos\theta\cos \theta + \cos^2\theta}$$
$$=\frac {1-\cos^2\theta + i\sin\theta\cos \theta}{1- \cos^2\theta}$$
$$=1+  i\cot\theta$$
