algebra precalculus - Writing the complex number $z = 1 - sin{alpha} + icos{alpha}$ in trigonometric form

Now I can't finish this problem:

Express the complex number $z = 1 - \sin{\alpha} + i\cos{\alpha}$ in trigonometric form, where $0 < \alpha < \frac{\pi}{2}$.

So the goal is to determine both $r$ and $\theta$ for the expression: $z = r(\cos{\theta} + i\sin{\theta})$

I've done this so far:

1. First of all I obtained $r = \sqrt{(1-\sin{\alpha})^2 + \cos^2{\alpha}} = \sqrt{1 + 2 \sin{\alpha} + \sin^2{\alpha} + \cos^2{\alpha}} = \sqrt{2(1 - \sin{\alpha})}$ (possible thanks to the condition over $\alpha$).




2. Now I tried to get $\theta = \arctan{\left(\frac{\cos{\alpha}}{1-\sin{\alpha}}\right)}$

And here it is where I get stuck... how to determine $\theta$ with such an expression?

I already know $0 < 1-\sin{\alpha} < 1$ and $0 < \cos{\alpha} < 1$ under the given conditions.

Any help will be appreciated. Thank you :)

P.S. I think (according to my search results here) there are no questions about this problem. I hope you won't mind if it is a duplicate.