Kathleen Rausch

2021-12-27

Why is $\mathrm{tan}\theta \approx \frac{1}{\frac{\pi }{2}-\theta }$ for $\theta$ close to $\frac{\pi }{2}$?
I wanted to see what the behaviour of the steep part of the $\mathrm{tan}$ curve was like, i.e. the behaviour of . So by thinking about a shift of the graph of to the left, I put some small (positive and negative) values of $\theta$ into my calculator for the function $\mathrm{tan}\left(\theta +\frac{\pi }{2}\right)$
$\mathrm{tan}\theta \approx \frac{1}{\frac{\pi }{2}-\theta }$ for $\theta$ close to $\frac{\pi }{2}$?
or, in more colloquial terms,
The steep part of $\mathrm{tan}x$ is just like the steep part of $\frac{1}{x}$
But why is this the case? I couldn't deduce it easily using the Maclaurin expansion of $\mathrm{tan}\left(x\right)$. Is there a more intuitive explanation? I couldn't think of any explanations analogous to those explaining small angle approximations.

amarantha41

$\mathrm{tan}\theta \equiv \mathrm{cot}\left(\frac{\pi }{2}-\theta \right)\equiv \frac{\mathrm{cos}\left(\frac{\pi }{2}-\theta \right)}{\mathrm{sin}\left(\frac{\pi }{2}-\theta \right)}$
Can you see it now?

Rita Miller

So, basically you would like to see that
$\underset{\theta \to {\left(\frac{\pi }{2}\right)}^{-}}{lim}\left(\left(\frac{\pi }{2}-\theta \right)\mathrm{tan}\theta \right)=1$
Or, letting $t\phantom{\rule{0.222em}{0ex}}=\frac{\pi }{2}-\theta$
$\underset{t\to {0}^{+}}{lim}\left(t\mathrm{tan}\left(\frac{\pi }{2}-t\right)\right)=\underset{t\to {0}^{+}}{lim}\left(\frac{t}{\mathrm{sin}t}\mathrm{cos}t\right)=1$
which is true.

Vasquez

Note that
$\mathrm{tan}\left(\theta \right)=\mathrm{cot}\left(\frac{\pi }{2}-\theta \right)$
And Maclaurean series for $\mathrm{cot}u$ where $u\to 0$ is:
$\mathrm{cot}u=\frac{1}{u}-\frac{u}{3}+O\left(u\right)$

Do you have a similar question?