Recently I came up with an algebra problem with a nice geometric representation. Basically, I would like to know what happens if we repeatedly circumscribe a rectangle by another rectangle which is rotated by $\alpha \in \left( 0, \frac {\pi} {4}\right)$ radians. Use this picture as reference:

In particular, do the resulting rectangles converge to a square?

It is rather easy to show that the rectangles do converge to a square if $\alpha$ is constant throughout the process. However, if we make it more general by defineing a sequence $\left(\alpha_n\right)_{n=1}^{\infty}$ of angles and use $\alpha_i$ in the $i$'th operation, then the answer seems to depend on the chosen sequence. So, for which sequences $\left(\alpha_n\right)_{n=1}^{\infty}$ do the rectangles converge to a square?

Algebraically this problem can be defined like this:

Define two real sequences by $A_0=a, B_0=b, a \neq b, a,b \in R_{\gt0}$ and $A_{n+1}=B_n\sin\alpha_n + A_n\cos\alpha_n, B_{n+1}=A_n\sin\alpha_n + B_n\cos\alpha_n \forall n \in N_{\gt 0}$, where $\alpha_i \in \left( 0, \frac {\pi} {4}\right) \forall i \in N_{\ge 0}$. Is it true that $\lim_{n \to \infty}\frac{A_n}{B_n}=1$?

I tried out a few sequences in C++ to notice some patterns. Interestingly, rectangles seem to converge to a square if and only if $\lim_{n \to \infty} \left( \sum_{i=0}^n \alpha_i \right) = \infty$. In particular, for $\alpha_n = \frac{1}{n}$ the convergence is really slow, however, it still seems to be converging.

Also, I believe that showing $\lim_{n \to \infty}\left(A_n-B_n\right)=1$ would be an even stronger result for this problem. Does such a replacement have any influence on the result?

For the record, I am still a high school student, so I have no idea how hard this problem might actually be. Any help would be highly appreciated.

P.S. This is my first question on the site, so please don't judge my wording and style too much. Feel free to ask questions if anything is unclear.