How to identify and classify singularities of $\frac{\sin(z)}{1-\tan(z)}$

I want to identify and classify the singularities of $$\frac{\sin(z)}{1-\tan(z)}$$. There are obvious singularities where $$\tan(z)=1$$. I have two problems:

1. I don't know how to classify $$z = \frac{1}{2}(n+1)\pi$$ where $$\tan$$ is not defined, are those points even singularities?
2. How do I find the Laurent series? I don't know if it can even be found directly. Of course, I know the power series for sin and tan, but there is division involved. Is there a smarter way to classify the singularities?

Note that $$\displaystyle \lim_{z \to \frac{(n+1)\pi}{2}} \frac{\sin z}{1-\tan z} = \lim_{z \to \frac{(n+1)\pi}{2}} \frac{\sin z \cos z}{\cos z-\sin z} = 0$$. So those points are removable singularities.

You can check that $$\displaystyle \tan z = 1 \iff z = \frac{\pi}{4} + n\pi, \, n\in\mathbb{Z}$$ and those points are simple poles. Hence Laurent series becomes $$\displaystyle \frac{a_{-1}}{z-w_n} + a_0 + a_1(z-w_n) + a_2(z-w_n)^2 + \cdots$$ where $$\displaystyle w_n = \frac{\pi}{4} + n\pi$$. This gives:

$$\displaystyle \frac{\sin z}{1-\tan z} = \frac{a_{-1}}{z-w_n} + a_0 + a_1(z-w_n) + a_2(z-w_n)^2 + \cdots$$

$$\displaystyle \sin z = (1-\tan z) \left( \frac{a_{-1}}{z-w_n} + a_0 + a_1(z-w_n) + a_2(z-w_n)^2 + \cdots \right)$$. Use this equality to find Laurent coefficients by considering Taylor series of $$\sin z$$ and $$1-\tan z$$ at $$\displaystyle z=w_n$$.

For example, let $$n=0$$ so $$w_0 =\dfrac{\pi}{4}$$

$$\displaystyle \sin z = \frac{1}{\sqrt{2}} + \frac{x-\frac{\pi}{4}}{\sqrt{2}} - \frac{\left( x-\frac{\pi}{4} \right)^2}{2\sqrt{2}} - \frac{\left( x-\frac{\pi}{4} \right)^3}{6\sqrt{2}} + \frac{\left( x-\frac{\pi}{4} \right)^4}{24\sqrt{2}} + \frac{\left( x-\frac{\pi}{4} \right)^5}{120\sqrt{2}} - \cdots$$

$$\displaystyle = \left( -2\left( x-\frac{\pi}{4} \right) - 2\left( x-\frac{\pi}{4} \right)^2 - \frac{8}{3}\left( x-\frac{\pi}{4} \right)^3 + \cdots \right) \left( \frac{a_{-1}}{z-\frac{\pi}{4}} + a_0 + a_1\left(z-\frac{\pi}{4}\right) + a_2\left(z-\frac{\pi}{4}\right)^2 + \cdots \right)$$

$$\displaystyle \implies -2a_{-1} = \frac{1}{\sqrt{2}}, \, -2a_0 -2a_{-1} = \frac{1}{\sqrt{2}}, \, -2a_1 -2a_0 -\frac{8}{3}a_{-1} = -\frac{1}{2\sqrt{2}}, \, \dots$$ by multiplying and comparing the terms of same degree.