Trignometric cardinality

On the interval $0<x<\frac{\pi}{2}$ , the functions $\cos(\cos(x))$ and $\sin(\sin(x))$ are both strictly increasing, and satisfy $0<\sin(\sin(x))<\cos(\cos(x))<1$

The function $\tan(\tan(x))$ takes all real values, and has infinitely many vertical asymptotes; in particular, $\tan(\tan(x))=0$ has infinitely many roots of the form $x=a_k=\tan^{-1} (k\pi)$ and $\tan(\tan(x))=1$ has infinitely many roots of the form $x=b_k=\tan^{-1} \left(k\pi + \frac{\pi}{4}\right)$ , where $k$ is an integer.

Since arctan is a strictly increasing function, we have $a_k<b_k<a_{k+1}<b_{k+1} \cdots$ , and $\begin{aligned} 0&<a_k<b_k<\frac{\pi}{2} \\ \tan(\tan(a_k))&=0 \\ \tan(\tan(b_k))&=1 \\ a_k<x<b_k &\Rightarrow 0<\tan(\tan(x))<1 \end{aligned}$

and $\tan(\tan(x))$ is continuous on $[a_k,b_k]$ .

There are therefore roots to both $\cos(\cos(x))=\tan(\tan(x))$ and $\sin(\sin(x))=\tan(\tan(x))$ in the interval $(a_k,b_k)$ .

This proves both sets have infinite cardinality, so $|A|=|B|$ because we can make a bijection between the two sets (like Hilbert's hotel, or the fact the cardinality of $\mathbb{Z}$ is the same as the cardinality of $\mathbb{N}$ ).