How would I prove $\frac{1}z=\operatorname{cis}(-\theta)$ given $z=\operatorname{cis}(\theta)$? I know $\operatorname{cis}(-\theta)$ equals $\cos(\theta)-i\sin(\theta)$
and $\frac{1}z=\frac{1}\cos+i\frac{1}\sin=\sec+\frac{\csc}i$
(please, correct me if I'm wrong.)
I just don't know how to relate these two now. Can anyone provide some guidance?
 A: It is not correct in general to say if $z=a+bi$ then $\frac 1 z = \frac 1 a + \frac 1 b i; $
rather, $\frac 1 z = \frac a {a^2+b^2} - \frac b {a^2+b^2}i. $ 
Note that $\cos^2\theta + \sin^2\theta = 1$.
Therefore, if $z=e^{i\theta}=\cos \theta + i \sin \theta$ then $\frac 1 z =e^{-i\theta}=\cos\theta-i\sin\theta.$
A: assuming you mean $\text{cis}(\theta)=e^{\theta i}=\cos\theta+i\sin\theta$ you can just write
$$\text{cis}(-\theta)=e^{-\theta i}=\cos(-\theta)+i\sin(-\theta)=\cos\theta-i\sin\theta\\
\frac{1}{z}=\frac{1}{\cos\theta+i\sin\theta}\\
=\frac{1}{\cos\theta+i\sin\theta}\cdot\frac{\cos\theta-i\sin\theta}{\cos\theta-i\sin\theta}\\
=\frac{\cos\theta-i\sin\theta}{\cos^2\theta-(i\sin\theta)^2}\\
=\frac{\cos\theta-i\sin\theta}{\cos^2\theta+\sin^2\theta}=\cos\theta-i\sin\theta=\text{cis}(-\theta)$$
where you just multiply by the conjugate to eliminate the complex number on below part, and use the facts that $i^2=-1$ and $\cos^2\theta+\sin^2\theta=1.$
P.s.: the step you took
$$\frac{1}{z}=\frac{1}{\cos\theta}+\frac{i}{\sin\theta}$$
is wrong because you can't generally say that $\frac{1}{a+bi}=\frac{1}{a}+\frac{i}{b}.$
