This question stemmed from asking, "if I can use rotation matrices to rotate any point in the x-y plane about the origin, then so long as a function is well-defined, the function is just a set of points, so if I rotate every point of (the graph of the) function anticlockwise by theta degrees about the $x$-axis, then this will give me the original graph but rotated." Pretty neat. So why not do it with $\tan(x)$? I'm mainly interested in rotating by 45 degrees anti about the origin, but any angle that gives an explicit answer would be interesting.
Initial working on paper attached below. Apologies for it being messy.
Rotating $\tan(x)$ anywhere between 45 degrees and -135 degrees anticlockwise about the origin will give a function, so setting aside the "obvious" $-\arctan(x)$, which is a rotation of $\tan(x)$ by 90 degrees anti about the origin, I would guess that, for all possible rotations that give a function, either all can be written as explicit functions, or none of them can. But I could be wrong about this. By the way, I'm not sure about the precise definitions of "closed form" and "explicit"- what is the difference? Does explicit = closed form but nonparametric? I think I want an explicit function. I want what Anixx wanted based on the comments of his/her question. So basically in the form "$y = f(x)$". So on page 2 of my paper working above, from the parametric equations, we get $$\tan( \frac{1}{\sqrt2} (x+y)) = \frac{1}{\sqrt2}(y-x),$$ but this is not in the form $y= f(x)$, which is what I want.
I also had a look here: and tried some graphs on WolframAlpha, but couldn't get anything similar for $tan(x)$ rotated by 45 degrees anticlockwise about the origin. Something like this isn't a million miles away, but for the function I'm after, the limit of the gradient of the function as x approaches 0 tends to -infinity. The gradient at $x=0$ of $-\arctan(x) -x$ is -2. So $-\arctan(x) - x$ would be like lhf's answer in the above link. "Similar" to what I want, but not exactly what I want. Based on the fact that Jason T. Miller gave an explicit formula using a complex analytic maps, on wolframalpha I tried something similar for tanz but couldn't get what I wanted. e.g. this is nowhere near close. I tried other stuff but got nowhere.
Back to this question, I wonder if the answer to my question is something not so pretty, like Jack D'Aurizio's comment,
"Essentially, if you want to express $t+\sin(t)$ as a function of $t−\sin(t)t$, you need a closed-form expression for the inverse function of $t−sint=t^3 /6−t^4 /24+t^5 /120−\ldots.$"
I'm not that familiar with the implicit function theorem either and I'm not sure how to or if I even should use it here.
If possible, I suspect the answer to this question is some combination of $x, \tan(x)$ and $\arctan(x)$.
Update: small progress: the function will, at best, be piece-wise (of 2 functions) because a "nice" function cannot have a derivative of -infinity for any value of x, and ours does for x=0. However, I do think our function is really a (continuous) function. It's just not differentiable (at x=0) and therefore not one nice function. Kind of like the piece-wise function $y = sqrt(-x) \quad for \ x<0,\quad y= -sqrt(x) \quad for \ x>=0$
Also, I believe our line is the reflection in the line y = x - and therefore the inverse - of $y = -x + arctan(x)$. Maybe multiplied by $\sqrt(2)$ , but that I can figure out later. I had a look at this method: https://www.quora.com/What-is-the-inverse-of-x-+-tanx
However, applying the method in the quora link above and assuming $y = -x + arctan(x)$, I get our function is f(x) = -x +1/x + C, so something must be off there.
Lastly, the derivative of our function is going to look something like $-1-1/x^2$, caring about the asymptotic appearance of the graph.
Assume a point $P_1=(x_1,y_1)$ on the $xOy$ plane. If we rotate $OP_1$ around $O$ in an angle $\phi$ (anticlockwise), then the new point ''say'' $P_2=(x_2,y_2)$ will have coordinates $$ x_2=x_1\cos\phi-y_1\sin\phi\textrm{ , }y_2=x_1\sin\phi+y_1\cos\phi.\tag 1 $$ Assume that $\phi=\frac{\pi}{4}$, then $$ x_2=x_1\frac{\sqrt{2}}{2}-y_1\frac{\sqrt{2}}{2}\textrm{ , }y_2=x_1\frac{\sqrt{2}}{2}+y_1\frac{\sqrt{2}}{2}, $$ since $\sin\left(\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$. Hence rotating $y=\tan x$ with angle $\phi=\frac{\pi}{4}$, we get ($y_1=\tan x_1$) $$ \sqrt{2}x_2=x_1-\tan x_1\textrm{ , }\sqrt{2}y_2=x_1+\tan x_1.\tag 2 $$ Hence upon solving with respect to $x_1$ and $\tan x_1$ we get $$ x_1=\sqrt{2}\frac{x_2+y_2}{2}\textrm{ , }\tan x_1=\sqrt{2}\frac{y_2-x_2}{2}.\tag 3 $$ Lastly substituting $x_1$ (of first equation above) to the second equation, we obtain $$ \tan\left(\frac{x_2+y_2}{\sqrt{2}}\right)=\frac{y_2-x_2}{\sqrt{2}}. $$ Hence the rotated curve of $y=\tan x$, is $$ \frac{y-x}{\sqrt{2}}=\tan\left(\frac{y+x}{\sqrt{2}}\right)\tag 4 $$ Hence $$ y=y(x)=x+\sqrt{2}L^{(-1)}\left(\sqrt{2}x\right),\tag 5 $$ where $$ L(x)=\arctan\left(x\right)-x.\tag 6 $$
...CONTINUING
About the questions.
1) We have clearly that $L(x)$ is from $\textbf{R}\rightarrow\textbf{R}$ and $''1-1''$. Hence invertable in $\textbf{R}$. Also $L(x)\in C^{(\infty)}\left(\textbf{R}\right)$ i.e. infinite times differentiatable in $\textbf{R}$. $$ y=x+\sqrt{2}L^{(-1)}\left(\frac{2x}{\sqrt{2}}\right)\Leftrightarrow\frac{y-x}{\sqrt{2}}=L^{(-1)}\left(\frac{2x}{\sqrt{2}}\right)\Leftrightarrow L\left(\frac{y-x}{\sqrt{2}}\right)=\frac{2x}{\sqrt{2}}\Leftrightarrow $$ $$ \arctan\left(\frac{y-x}{\sqrt{2}}\right)-\frac{y-x}{\sqrt{2}}=\frac{2x}{\sqrt{2}}\Leftrightarrow\arctan\left(\frac{y-x}{\sqrt{2}}\right)=\frac{y+x}{\sqrt{2}}\Leftrightarrow $$ $$ \frac{y-x}{\sqrt{2}}=\tan\left(\frac{x+y}{\sqrt{2}}\right). $$ 2) If we set $S(x)=L^{(-1)}(x)$, then $$ \int^{S(x)}_{0}\frac{t^2}{t^2+1}dt=-x.\tag 7 $$ and $$ y=x+\sqrt{2}S\left(\sqrt{2}x\right).\tag 8 $$ These all I can find. About the inversion it may work the Lambert function $W(z)$, $W(z)e^{W(z)}=z$. But I don't think so.