 Kristina Mcclain

2022-01-26

Why ${\mathrm{tan}}^{-2}\left(x\right)=\frac{1}{{\mathrm{tan}}^{2}\left(x\right)}$? Kyler Jacobson

Expert

As Arturo Magidin said in the comments, it is a matter of notation. Usually, ${\mathrm{sin}}^{n}\left(x\right),{\mathrm{cos}}^{n}\left(x\right),{\mathrm{tan}}^{n}\left(x\right)$ denote ${\left(\mathrm{sin}\left(x\right)\right)}^{n},{\left(\mathrm{cos}\left(x\right)\right)}^{n},{\left(\mathrm{tan}\left(x\right)\right)}^{n}$ respectively, when n=1,2,…. Unfortunately, due to various historical reasons, when n=−1, the notation ${\mathrm{sin}}^{-1}\left(x\right),{\mathrm{cos}}^{-1}\left(x\right),{\mathrm{tan}}^{-1}\left(x\right)$ do not at all mean 1 divided by the respective functions, but rather represent the inverse function.
Your confusion probably comes from your teacher/textbook using ${\mathrm{tan}}^{-2}\left(x\right)$ to mean ${\left(\mathrm{tan}\left(x\right)\right)}^{-2}$ instead of ${\left({\mathrm{tan}}^{-1}\left(x\right)\right)}^{2}$ or ${\mathrm{tan}}^{-1}\left({\mathrm{tan}}^{-1}\left(x\right)\right)$. This confusing notation unfortunately rather common. The good news is that there's nothing conceptual you're missing, it's just an instance of particularly confusing notation.

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