Why is \tan \theta \approx \frac{1}{\frac{\pi}{2}-\theta}

Kathleen Rausch

Kathleen Rausch

Answered question


Why is tanθ1π2θ for θ close to π2?
I wanted to see what the behaviour of the steep part of the tan curve was like, i.e. the behaviour of tan(x) as x(π2). So by thinking about a shift of the graph of tan(x) byπ2 to the left, I put some small (positive and negative) values of θ into my calculator for the function tan(θ+π2)
tanθ1π2θ for θ close to π2?
or, in more colloquial terms,
The steep part of tanx is just like the steep part of 1x
But why is this the case? I couldn't deduce it easily using the Maclaurin expansion of tan(x). Is there a more intuitive explanation? I couldn't think of any explanations analogous to those explaining small angle approximations.

Answer & Explanation



Beginner2021-12-28Added 38 answers

Can you see it now?
Rita Miller

Rita Miller

Beginner2021-12-29Added 28 answers

So, basically you would like to see that
Or, letting t=π2θ
which is true.


Expert2022-01-08Added 669 answers

Note that
And Maclaurean series for cotu where u0 is:

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