Why is the inverse tangent function not equivalent to the reciprocal of the tangent function? I know that
$$
{\tan}^2\theta = {\tan}\theta \cdot {\tan}\theta
$$
So I guess the superscript on a trigonometric function is just like a normal superscript:
$$
{\tan}^x\theta = {({\tan}\theta)}^{x}
$$
Then why isn't this true?
$$
{\tan}^{-1}\theta = \dfrac{1}{{\tan}\theta}
$$
 A: The reason that it isn't true is that, regrettably, the notation is not consistent. For this reason, many people avoid using $\tan^{-1}$ and use $\arctan$ instead, and so on for the other trigonometric functions. That said, $\tan^{-1}$ is logical notation, and such notation as $\tan^2$ is illogical. However, the weight of tradition and the simple convenience of the latter notation ensures its survival, and we will probably always be using it.
A: It's better to interpret $\tan^{-1}$ as an inverse operator.
First of all, for a bijective (or one-to-one correspondence) function $f(x)$.  
We have
$$f^{-1}(f(x)) \equiv (f^{-1} \circ f) (x) \equiv x \\$$
$$f(f^{-1}(x)) \equiv (f \circ f^{-1}) (x) \equiv x \\$$
May see more on function composition.
Secondly, for an invertible (non-singular) matrix $\boldsymbol{A}$:
$$\boldsymbol{Ax=b} \iff \boldsymbol{x=A}^{-1}\boldsymbol{b}$$
We seldom write $$\boldsymbol{x=\frac{b}{A}}$$
though we sometimes write a Jacobian matrix as$$\boldsymbol{J}=\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{x}} \\$$
Thirdly, Fourier Transformation and its inverse transformation:
$$\mathcal{F}[f(x)]=F(\omega) \iff \mathcal{F}^{-1}[F(\omega)]=f(x) \\$$
Of course, some operations are many-to-one:
For example,
$$\theta=\sin^{-1} x \implies x=\sin \theta$$
but $$y=\sin \phi \implies \phi=n\pi+(-1)^{n} \sin^{-1} y$$
in which $\sin^{-1}$ refers to the principal value and $n\in \mathbb{Z}$.
Also derivative and its antiderivative,
if
$$\frac{dF(x)}{dx}=f(x)$$
then $$\int f(x) dx = F(x)+C$$
A: Depending on context, a raised expression on the right of a function name but before the parenthesized argument (e.g. $f^3(x)$) can be interpreted in two ways:


*

*as an exponent, $f^3(x)=f(x)\cdot f(x)\cdot f(x)$,

*as an iteration count $f^3(x)=f(f(f(x)))$.
Also note that with parenthesis around, you usually denote a derivative,
$$f^{(3)}(x)=f'''(x).$$
With a $-1$, you can have


*

*the reciprocal, also called multiplicative inverse, $f^{-1}(x)=\dfrac1{f(x)}$,

*the (compositional) inverse, such that $f(f^{-1}(x))=x=f^{-1}(f(x)).$


(Then $f^{(-1)}$ could denote an antiderivative, but I have never seen that.)
On another hand, wrapping the function with parenthesis usually excludes any ambiguity: $(f(x))^2$ and $(f(x))^{-1}$ are always understood as powers.
The lesson is that you must make sure to be understood, and make things explicit if the context doesn't suffice. For some functions, you have the option to use established names
$$\arctan(x)\leftrightarrow\cot(x).$$
A: Superscripts above a function usually refer to one of two things: repeated $composition$ or repeated $pointwise$ $multiplication$. Unfortunately, as in trigonometry, there are many times where both uses are needed at the same time, leading to some confusion. Sometimes, repeated composition is referred to by $f^{\circ n}$ to reduce this confusion, although admittedly, I have not seen $\tan^{\circ-1}(x)$ until having typed it right now.
