Why tan−2(x)=1tan2(x)?

As Arturo Magidin said in the comments, it is a matter of notation. Usually, ${\displaystyle {\sin }^n\left(x\right),{\cos }^n\left(x\right),{\tan }^n\left(x\right)}$ denote ${\displaystyle {\left(\sin \left(x\right)\right)}^n,{\left(\cos \left(x\right)\right)}^n,{\left(\tan \left(x\right)\right)}^n}$ respectively, when n=1,2,…. Unfortunately, due to various historical reasons, when n=−1, the notation ${\displaystyle {\sin }^{-1}\left(x\right),{\cos }^{-1}\left(x\right),{\tan }^{-1}\left(x\right)}$ 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 ${\displaystyle {\tan }^{-2}\left(x\right)}$ to mean ${\displaystyle {\left(\tan \left(x\right)\right)}^{-2}}$ instead of ${\displaystyle {\left({\tan }^{-1}\left(x\right)\right)}^2}$ or ${\displaystyle {\tan }^{-1}\left({\tan }^{-1}\left(x\right)\right)}$. 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.