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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?

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The power series for cotangent (for $x$ in radians) is $$\cot x = \frac{1}{x} - \frac{1}{3}x - \frac{1}{45}x^3 - \frac{2}{945}x^5 - \cdots \approx \frac{1}{x} \; \text{for $x$ small}$$

So,

$$\tan\left( 90^\circ - \frac{1^\circ}{10^n} \right) = \cot\frac{1^\circ}{10^n} = \cot\frac{\pi/180}{10^n} \approx \frac{10^n}{\pi/180} = \frac{180}{\pi}\times 10^n$$

Congratulations for noticing the pattern on your calculator!

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We have $\tan(\frac{\pi}{2}-x)=\frac{\cos(x)}{\sin(x)}$. But $\cos$ is continuous at $0$, so $\cos(x)\approx \cos(0)=1$ for $x$ close enough to $0$. As well, $\sin$ is differentiable at $0$, so $\sin(x)$ can be approximated by its tangent at $0$, which is $x$. Thus, $\tan(\frac{\pi}{2}-x)\approx\frac{1}{x}$ when $x$ is close enough to $0$

(You can have an estimation of the error made by using Taylor's estimation).

Note: as always in maths, all the angles above are in radians.

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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 $$\tan\left(90^\circ-\frac{1^\circ}{10^n}\right)\approx\frac{1}{\frac{1}{10^n}\times\frac{\pi}{180}}=\frac{180}{\pi}\times 10^n.$$

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The tangent of $x$ degrees is $\tan(kx)$ where $k=\dfrac{\pi}{180}$. Let $\varepsilon=10^{-n}$. Your calculator is telling you that $\tan(k(90-\varepsilon))\approx\dfrac 1{k\varepsilon}$, in other words, $k\varepsilon\tan(\dfrac{\pi}2-k\varepsilon)\approx 1$. Note that $k\varepsilon\tan(\dfrac{\pi}2-k\varepsilon)=k\varepsilon\cot(k\varepsilon)=\dfrac{k\epsilon}{\sin(k\varepsilon)}\cdot\cos(k\varepsilon)\rightarrow 1$ as $\varepsilon\rightarrow 0$.

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