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 $\mathrm{tan}\left(x\right)\text{}as\text{}x\to {\left(\frac{\pi}{2}\right)}^{-}$ . So by thinking about a shift of the graph of $\mathrm{tan}\left(x\right)\text{}by\frac{\pi}{2}$ to the left, I put some small (positive and negative) values of $\theta$ into my calculator for the function $\mathrm{tan}(\theta +\frac{\pi}{2})$

$\mathrm{tan}\theta \approx \frac{1}{\frac{\pi}{2}-\theta}$ 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.

I wanted to see what the behaviour of the steep part of the

or, in more colloquial terms,

The steep part of

But why is this the case? I couldn't deduce it easily using the Maclaurin expansion of

amarantha41

Beginner2021-12-28Added 38 answers

Can you see it now?

Rita Miller

Beginner2021-12-29Added 28 answers

So, basically you would like to see that

$\underset{\theta \to {\left(\frac{\pi}{2}\right)}^{-}}{lim}\left((\frac{\pi}{2}-\theta )\mathrm{tan}\theta \right)=1$

Or, letting$t{\textstyle \phantom{\rule{0.222em}{0ex}}}=\frac{\pi}{2}-\theta$

$\underset{t\to {0}^{+}}{lim}\left(t\mathrm{tan}(\frac{\pi}{2}-t)\right)=\underset{t\to {0}^{+}}{lim}\left(\frac{t}{\mathrm{sin}t}\mathrm{cos}t\right)=1$

which is true.

Or, letting

which is true.

Vasquez

Expert2022-01-08Added 669 answers

Note that

And Maclaurean series for

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