What is the value of $\tan (\frac{\pi}{2} - \epsilon)$?

Question

$$\tan(\frac{\pi}{2}-\epsilon)$$

By epsilon I mean an infinitesimal.

Answer 1

For any $\epsilon$, $$\tan\left(\frac{\pi}{2}-\epsilon\right)=\frac{\sin\left(\frac{\pi}{2}-\epsilon\right)}{\cos\left(\frac{\pi}{2}-\epsilon\right)}=\frac{\cos \epsilon}{\sin \epsilon}=\cot \epsilon.$$ For small $\epsilon$, this can be approximated as $\frac{1}{\epsilon}-\frac{\epsilon}{3},$ since $\cot\epsilon\approx\frac{1}{\epsilon}\frac{1-\epsilon^2/2}{1-\epsilon^2/6}\approx\frac{1}{\epsilon}\left(1-\frac{\epsilon^2}{2}+\frac{\epsilon^2}{6}\right).$

Answer 2

You mean $$\tan \left( \frac{\pi}{2} - \epsilon \right) \sim \frac{1}{\epsilon} \quad \text{when } \epsilon \rightarrow 0$$

Explanation

Used $$\tan(x) = \frac{\sin(x)}{\cos(x)} \approx \frac{1}{\frac{\pi}{2}-x}$$

by doing a taylor expansion about $x=\frac{\pi}{2}$ separately for the sine and cosine.

Then use $x=\frac{\pi}{2}-\epsilon$ to get your answer.

If used a higher order taylor series you also get

$$\tan\left( \frac{\pi}{2}-\epsilon \right) \approx \frac{\epsilon^2-2}{2 \epsilon} \approx \frac{3 ( \epsilon^2-2)}{\epsilon (6-\epsilon^2)} $$