$\arctan x$ vs $1/\tan x$ Why is $\arctan x = (\tan x)^{-1} = \tan^{-1}x$ but is not equal to $1/\tan x$, even though $\tan^{2}x = (\tan x) \cdot (\tan x)$?
Is there not another way to write $1/\tan x$?
 A: No one invented all of mathematical notation at once and consistently, and to answer your “Why?” isn’t easy. Over time, mathematicians proposed or invented notations that they felt worked and used them in publications. Sometimes conflicting notations caught on and became conventional in modern mathematics. Most of the time, context makes things clear, and we eventually don’t even notice the inconsistencies.
The uses of superscripted numbers like $^{-1}$ and $^2$ to mean different things in different contexts are just one example. Another is simple grouping and juxtaposition: $f(x+y)$ rarely means $f\times (x+y)$, but $2(x+y)$ usually does mean $2\times(x+y)$, but $z(x+y)$, well, it likely depends (on the context).
Another example: What does $\left({1\  2\  3\atop 2\ 3\ 1}\right)$ mean? Is it an element of the permutation group $S_3$, or is it a $2$ by $3$ matrix? It depends on the context.
If you want to go down a rabbit hole with this, check out “A history of mathematical notations,” by Florian Cajori, available here and elsewhere.
A: The notation $\tan^2(x)$ "by default" denotes the square of the tangent because it is exceptional to meet the iterated function $\tan(\tan(x))$ that the exponent $2$ might denote.
On the opposite, $\tan^{-1}(x)$ might as well denote the functional inverse ($\arctan(x)$) or the multiplicative inverse ($\cot(x)$), because the two functions are frequently used. I guess that the first interpretation is more frequent, but without guarantee.
The moral of the story: make sure that by context what you mean is unambiguous, or state it explicitly.
A: Strangely, nobody seems to have answered the second part of your question:

Is there not another way to write  $1/\tan x$?

The answer is yes:  the multiplicative inverse of the tangent function is called the cotangent function, and is denoted $\cot x$.  You can also write this as $(\tan x)^{-1}$, if you prefer.  Notice that the parentheses are playing an important role, here!
You may notice that in the sentence above I used the phrase "multiplicative inverse" to describe the relationship between $\tan$ and $\cot$.  This phrase means something different from "inverse function", which describes the relationship between $\tan$ and $\tan^{-1}$, and I think this distinction is at the heart of your question, so let me take a minute to explain the underlying issue.
The set of all functions (or, if you prefer, all functions defined on a specific interval) is not just a set of individual things, like buttons on a calculator; there are relationships among them, and operations that can be performed on them.  For example, you can add two functions $f$ and $g$ together, to get a new function $f + g$.  You can also multiply two functions together to get a new function $fg$.  And there is a third operation:  you can compose two functions to get $f\circ g$, defined by the rule $(f\circ g) (x) = f(g(x))$.
With respect to each of these operations, you can ask some important questions:

*

*Is there an identity function with respect to this operation?

*If there is an identity function for a particular operations, do functions have inverses with respect to that operation?

Let's consider how we answer these questions for the three operations I've mentioned: addition, multiplication, and composition.
1.  Function addition:
An "additive identity" would be a function $k$ with the property that for any $f$, we would have $f + k = f$. Because of how function addition is defined, this is equivalent to $f(x) + k(x) = f(x)$, which in turn is equivalent to $k(x) = 0$.  So the constant function $k(x) = 0$ is the additive identity function.
Now what about additive inverses?  The question we ask is, given any function $f$, is there a "counterpart" function $g$ with the property that $f + g = k$, where $k$ is the additive identity function we just found?  The answer is yes:  we define $g$ by $g(x) = -f(x)$.  Then $g$ is the additive inverse of $f$.
2.  Function multiplication:
A "multiplicative identity" would be a function $i$ with the property that for any $f$, we would have $f \cdot i = f$. Because of how function multiplication is defined, this is equivalent to $f(x) \cdot i(x) = f(x)$, which in turn is equivalent to $i(x) = 1$.  So the constant function $i(x) = 1$ is the multiplicative identity function.
Now what about multiplicative inverses?  The question we ask is, given any function $f$, is there a "counterpart" function $g$ with the property that $f \cdot g = i$, where $i$ is the multiplicative identity function we just found?  The answer is yes, provided that $f(x)$ is always nonzero:  we define $g$ by $g(x) = \frac{1}{f(x)}$.  Then $g$ is the multiplicative inverse of $f$.  (If $f(x) = 0$ for one or more values of $x$ then $g$ would only be defined on a restricted domain; in that case we can say that $$g(x) = \frac{1}{f(x)}$ is a partial multiplicative inverse.)
3.  Function composition:
A "compositional identity" would be a function $e$ with the property that for any $f$, we would have $f \circ e= f$ and $e \circ f = f$.  (We need both requirements here because composition, unlike multiplication and addition, is not commutative). This is equivalent to requiring $f(e(x)) = f(x)$ and $e(f(x)) = f(x)$.  It is not too difficult to see that the function $e(x) = x$ has both of these properties, and in fact is the only function with both of these properties.  So the constant function $e(x) = x$ is the compositional identity function.   (People usually just call it the "identity function", and drop the word "compositional", but I'm going to keep it here because the distinction between function multiplication and function composition is at the core of your question.)
Now what about compositional inverses?  We want to know if, for any function $f$, there exists some $g$ with the property that $f \circ g = e$ and $g \circ f = e$; you might be better able to recognize these requirements in the form $f(g(x)) = x$ and $g(f(x)) = x$.  Such a function might not exist -- in particular, the second requirement requires that $f$ be a one-to-one function (you may recognize this as the "horizontal line test").  If $f$ is not invertible, we may be able to restrict the domain of $f$ to one in which it is one-to-one, in which case we can define a compositional inverse function, which we denote $f^{-1}(x)$.
Notice what has happened:  for each operation, there is an identity function, and a different kind of inverse function.

*

*For addition, the identity is $k(x) = 0$, and the additive inverse of $f(x)$ is $-f(x)$.

*For multiplication, the identity is $i(x) = 1$, and the multiplicative inverse of $f(x)$ is $\frac{1}{f(x)}$ (where defined).

*For composition, the identity is $e(x) = x$, and the composition inverse of $f(x)$ is $f^{-1}(x)$ (if defined).

Let's take a specific example, related to the one you asked about.  With $f(x) = \sin x$,

*

*the additive inverse is $-\sin x$

*the multiplicative inverse is $\frac{1}{\sin x}$, which is called the cosecant function, denoted $\csc x$

*the compositional inverse is $\sin^{-1}(x)$, which is also denoted $\arcsin x$, and is called the arcsine function.

Confusion between the two operations of "function multiplication" and "function composition" is commonplace and, to some extent, natural.  We use almost the same notation for both (compare $f \cdot g$ and $f \circ g$), and the notations $f^{-1}(x)$ and $(f(x))^{-1}$ are almost identical. I think at a deeper level the confusion between multiplication and composition also underlies common errors, like writing
$$\sqrt{x + y} = \sqrt{x} + \sqrt{y}$$
and other versions of what is called the "Freshman's Dream".  At a formal level, the error here is thinking that
$f \circ (g + h) = f \circ g + f \circ h$, i.e. believing that composition, like multiplication, distributes across addition.
