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

Question

If tan^-1(x) can't be 1/tanx then why can tan^-2(x) =1/tanx Actually I know than tan^-1(x) is the inverse trigonometry and it is equal to tan($\theta)$ so why don't tan^-2(x) is equal to tan^2($\theta)$.

Answer 1

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