Kristina Mcclain

Answered

2022-01-26

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

Answer & Explanation

Kyler Jacobson

Expert

2022-01-27Added 8 answers

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.

Your confusion probably comes from your teacher/textbook using

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