Determine the argument of the complex $Z$

Question

Determine the argument of the complex $$Z= \frac{1+\cos (8\theta) + i \sin (8\theta)}{\cos ^2 (\theta) (1- \tan ^2 (\theta) + 2 i \tan (\theta))}$$

Attempt: I realized that I can convert things from $ i \tan x $ to $(i\tan x+1)^2$. Expanding $ \tan x = \frac{\sin x}{\cos x} $, we are left with $e^{-2ix}+e^{6ix}$. I don't know how to simplify more than that

Answer 1

Since $e^{-2ix}$ and $e^{6ix}$ are of equal magnitude, they form two adjacent sides of a rhombus and the sum's argument is the average of those of its components: $\frac{-2x+6x}2=2x$.