Why does the tangent of numbers very close to $\frac{\pi}{2}$ resemble the number of degrees in a radian? Testing with my calculator in degree mode, I have found the following to be true:
$$\tan \left(90 - \frac{1}{10^n}\right) \approx \frac{180}{\pi} \times 10^n, n \in \mathbb{N}$$
Why is this? What is the proof or explanation?

First, note that by standard trigonometric identities,
$$\tan\left(90^\circ-\frac{1^\circ}{10^n}\right)= \frac{\sin\left(90^\circ-\frac{1^\circ}{10^n}\right)}{\cos\left(90^\circ-\frac{1^\circ}{10^n}\right)}=\frac{\cos\left(\frac{1^\circ}{10^n}\right)}{\sin\left(\frac{1^\circ}{10^n}\right)}=\frac{\cos\left(\frac{1}{10^n}\times\frac{\pi}{180}\right)}{\sin\left(\frac{1}{10^n}\times\frac{\pi}{180}\right)}.$$
Also, the power series for $\sin$ and $\cos$ are
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$
$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$$
so that, for small $x$, we have
$$\sin(x)\approx x\hphantom{,}$$
$$\cos(x)\approx 1,$$
and therefore
$$\tan\left(90^\circ-\frac{1^\circ}{10^n}\right)\approx\frac{1}{\frac{1}{10^n}\times\frac{\pi}{180}}=\frac{180}{\pi}\times 10^n.$$
