Why is tan⁡θ≈1π2−θ 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)tan⁡θ≈1π2−θ for θ close to π2?or, in more colloquial terms,The steep part of tan⁡x is just like the steep part of 1xBut 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.

Question

Why is tan⁡θ≈1π2−θ 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)tan⁡θ≈1π2−θ for θ close to π2?or, in more colloquial terms,The steep part of tan⁡x is just like the steep part of  1xBut why is this the case? I couldn&#039;t deduce it easily using the Maclaurin expansion of  tan⁡(x).  Is there a more intuitive explanation? I couldn&#039;t think of any explanations analogous to those explaining small angle approximations.

amarantha41 · Accepted Answer

tan⁡θ≡cot⁡(π2−θ)≡cos⁡(π2−θ)sin⁡(π2−θ)  Can you see it now?

Rita Miller · Accepted Answer

So, basically you would like to see that  limθ→(π2)−((π2−θ)tan⁡θ)=1  Or, letting t=π2−θ  limt→0+(ttan⁡(π2−t))=limt→0+(tsin⁡tcos⁡t)=1  which is true.

Vasquez · Accepted Answer

Note that tan⁡(θ)=cot⁡(π2−θ) And Maclaurean series for cot⁡u where u→0 is: cot⁡u=1u−u3+O(u)

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

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