Relativistic transformation of $c^2/v$ Everybody knows the relativistic transformation of a velocity $v$, here in one dimension,
$$v'=\frac{v+w}{1+\frac{w}{c^2} v}.$$
But then this implies that
$$\frac{c^2}{v'} = \frac{\frac{c^2}{v}+w}{1+\frac{w}{c^2}\frac{c^2}{v}},$$
i.e. the same transformation law for $v$ and for $c^2/v$, to hammer the point home!
I had never noticed that before. What is strange is that for a speed $|v|<c$, the speed $\frac{c^2}{|v|}>c$ cannot be that of any particle. It could be a group velocity but that does not shed any light for me.
It can't be just a coincidence, can it? What is the physical meaning of this "reciprocity"?
In order to connect more closely with @robphy's answer, I could have formulated my question as follow.
Given two inertial observers Bob and Carol assigning respective coordinates $(x^0, x^1)$ and $(y^0, y^1)$ to the same event, the transformation mapping a velocity with respect to Bob to a velocity with respect to Carol also maps (i) the "slope" $x^1/x^0$ onto $y^1/y^0$, and (ii) the slope $x^0/x^1$ onto $y^0/y^1$. This is natural for (i) as this slope can transparently be interpreted as the velocity of a signal which was at position $x^1=0$ at time $x^0=0$ but it comes as a surprise for (ii).
 A: Well, for what this is worth:
Fix $w$ and put $f(v)=(v+w)/(1+vw)$. Think of $f$ as a map from ${\mathbb R}^+$ to itself. 
(Here ${\mathbb R}^+$ is the positive real numbers.  $f$ is also defined for negative reals, but never mind that now.)
What you are saying is equivalent to:

$f$ commutes with the taking of inverses.

We can identify ${\mathbb R}^+$ with ${\mathbb R}$ by applying the log function in one direction and the exp function in the other.  These maps are homomorphisms and so take inverses to negatives.  After this identification, we have a map $\hat{f}:{\mathbb R}\rightarrow {\mathbb R}$ given explicitly by $x\mapsto \log(f(\exp(x)))$.  Under this identification, and accounting for what's been said about homomorphsims, what you are saying is that
 $\hat{f}$ commutes with multiplication by $-1$

or in other words 

$\hat{f}$ is an odd function.

This is easily confirmed by a direct calculation and/or by an appeal to the calculation in your post.
I'm not sure this adds anything, other than to point out that your observation comes down to the fact that a particular function is odd.  Odd functions are not all that rare, and it's not always considered odd that they crop up from time to time, so maybe that's all there is to it.
A: Rephrasing a bit, the relativistic addition law says that if $v_1$ and $v_2$ sum to $v$, which we'll write as $v_1 \oplus v_2 = v$, then their "boost angles" (or rapidities) add directly. That is,
$$\tanh \phi_i = v_i, \quad \tanh \phi = v, \quad \phi_1 + \phi_2 = \phi.$$
Your observation is that $1/v_1 \oplus v_2 = 1/v$ as well, which is equivalent to showing that
$$\text{arctanh}(a) - \text{arctanh}(1/a) = \text{constant}$$
for all $a \in (-1, 1)$. And this turns out to be true, so maybe your result just boils down to a coincidental property of arctanh.
I've tried to come up with a physical explanation. Having $|v| > 1$ corresponds to having a complex 'boost angle' $\phi$. I don't know what that means, and it's extra confusing because boost angles are already 'imaginary' generalisations of ordinary rotation angles (i.e. they are what you get if you rotate by an imaginary angle). So maybe $|v| > 1$ wraps it back around to being a normal rotation, but I can't see it. 
A: Under a boost by relative-spatial-velocity $w=v_{CB}=\tanh\omega$, 
Bob's 4-velocity vector is transformed into Carol's 4-velocity vector.
Call $\omega$ the relative-rapidity between the [timelike] 4-velocities.
The [spatial-]velocity transformation 
$$v_{CA}=\frac{v_{BA}+v_{CB}}{1+v_{BA}v_{CB}}$$
provides her spatial-velocity [the slope of Carol's 4-velocity vector]  $v'=v_{CA}=\tanh(\theta+\omega)$
in terms of the Bob's spatial-velocity $v=v_{BA}=\tanh\theta$ and the relative spatial velocity $w=v_{CB}=\tanh\omega$.
That boost also transforms 
Bob's "space axis" [with slope $1/v_{BA}$] into Carol's "space axis" [with slope $1/v_{CA}$]. (This is true because, in Minkowski spacetime geometry, two directions are perpendicular in that geometry when the product of the slopes is +1... to be contrasted with the Euclidean case where the product is $-1$. Note that by taking the product of the two formulas in the OP the right side simplifies.)
Although, technically speaking, the angle between spacelike lines isn't the rapidity... these spacelike lines are special because they are correspondingly perpendicular to the 4-velocities in the problem (that of Alice [not shown] and of Bob and Carol) and coplanar with all of them.


UPDATE:

Here is an algebraic calculation to support the spacetime diagram above.
I use signature $(+,-)$, where my first component is the time component.
Let $\hat V=\left(\begin{array}{c} \cosh{B} \\ \sinh{B} \end{array}\right)$ be Bob's 4-velocity (a timelike vector).
His spatial velocity $v=\rm{slope}=\frac{\rm spatial\ component}{\rm temporal\ component}=\frac{\sinh B}{\cosh B}=\tanh{B}$.
Let $M=\gamma\left(\begin{array}{cc} 1 & \beta \\ \beta & 1 \end{array}\right)$ be a boost that maps Bob's 4-vector to Carol's 4-vector. 
(The boost velocity $\beta$ corresponds to $w$ in the OP.)
I am intentionally mixing notations to distinguish the boost components from the 4-velocity components. 
So, we get Carol's 4-velocity 
\begin{align}
\hat V'&=M\hat V\\
\hat V'&=
\gamma\left(\begin{array}{cc} 1 & \beta \\ \beta & 1 \end{array}\right)
\left(\begin{array}{c} \cosh{B} \\ \sinh{B} \end{array}\right)\\
&=
\gamma\left(\begin{array}{c} \cosh{B} + \beta\sinh{B} \\ \beta\cosh{B}+\sinh{B} \end{array}\right) = \left(\begin{array}{c} \cosh{C} \\ \sinh{C} \end{array}\right)\\
\end{align}
$\hat V'$ has slope $$v'=\tanh{C}=\frac{\beta\cosh{B}+\sinh{B}}{\cosh{B} + \beta\sinh{B}}
=\frac{\beta+\tanh{B}}{1+\beta\tanh{B}}=\frac{\beta+v}{1+\beta v},$$
the velocity transformation (the OP's eq. 1).

Next...
Let $\hat V_{\perp}=\left(\begin{array}{c} \sinh{B} \\ \cosh{B} \end{array}\right)$ be Bob's unit-x vector (a spacelike vector orthogonal to his 4-velocity $\hat V$). It has slope $v_{\perp}=\frac{\cosh B}{\sinh B}=\coth B=\frac{1}{v}$.
The boost will map this vector to Carol's unit-x vector $\hat V_{\perp}'$ (a spacelike vector orthogonal to Carol's 4-velocity $\hat V'$).
So,
\begin{align}
\hat V_{\perp}'&=M\hat V_{\perp}\\
\hat V_{\perp}'&=
\gamma\left(\begin{array}{cc} 1 & \beta \\ \beta & 1 \end{array}\right)
\left(\begin{array}{c} \sinh{B} \\ \cosh{B} \end{array}\right)\\
&=
\gamma\left(\begin{array}{c} \sinh{B} + \beta\cosh{B} \\ \beta\sinh{B}+\cosh{B} \end{array}\right)
\end{align}
$\hat V_{\perp}'$ has slope $$v_{\perp}'=\frac{\beta\sinh{B}+\cosh{B}}{\sinh{B} + \beta\cosh{B}}
=\frac{\beta+\coth{B}}{1+\beta\coth{B}}=\frac{\beta+v_{\perp}}{1+\beta v_{\perp}}=\frac{\beta+(\frac{1}{v})}{1+\beta (\frac{1}{v})},$$
call it the spacelike-slope transformation (the OP's eq. 2).
For completeness, we should show that $v_{\perp}'=\frac{1}{v'}$.
\begin{align}
v_{\perp}'
&\stackrel{?}{=}\frac{1}{v'}\\
&\stackrel{?}{=}\frac{1+\beta v}{\beta + v}\\
&\stackrel{\surd}{=}\frac{(\frac{1}{v})+\beta }{\beta (\frac{1}{v}) +1}\\
\end{align}
...and we are done.
In my opinion, the PHYSICAL interpretation is that it is the slope of the observer's X-axis. (One can probably find other interpretations... but at the root of it... it is this slope.) 

UPDATE #2 (in response to the OP's update and comments):
In a Lorentz boost, we can see how the 4-velocity and spatial-axis of an observer transform in a complementary way by viewing a "light-clock diamond" on "rotated graph paper".
This is taken from a blog entry I contributed to at
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/
This is based on my recent paper (“Relativity on Rotated Graph Paper”, Am. J. Phys. 84, 344 (2016); http://dx.doi.org/10.1119/1.4943251 ).

On these diagrams, time runs upwards.
Light-Clock Diamonds are traced out by the spacetime paths of light-signals in a longitudinal light-clock. The worldlines of the mirrors of the light-clock are not shown, but are implied by the reflection events. In each tick of the light-clock, these events are "spatial" according to the observer carrying that light-clock. That is, that direction is Minkowski-perpendicular to that observer's 4-velocity. Visually, this means that the diagonals of a light-clock diamond are Minkowski-perpendicular to each other.
Under a boost, Alice's light-clock diamond must transform into Bob's light-clock diamond, which must have edges parallel to the light-cone (to preserve the speed of light) and must have area preserved (since the Lorentz boost has determinant equal to one). [The area turns out to be the square-interval of OF.]
Geometrically, the tip of the 4-velocity travels along the unit-hyperbola centered at the origin. [This ensures that the area of the diamond is preserved]. At the intersection point, the tangent to the hyperbola is Minkowski-perpendicular to the 4-velocity [a unit radius vector]. This tangent is parallel to the spacelike-diagonal of the light-clock diamond.
Thus, diagonal YZ is Minkowski-perpendicular to diagonal OF.
A: Link : About de Broglie relations, what exactly is E? Its energy of what?
In my answer in above link is contained the 3-dimensional version of this result. It's not strange neither funny. We may call it "the law of transformation of phase velocities" in analogy to "the law of transformation of particle velocities". Put the accompanied particle aside for a while : in 1-dimension a phase $\:\phi=\omega t\!-\!kx\:$ in a frame $\:\mathrm{S}\:$ propagating with speed $\:\mathrm{w}=\omega/k\:$ is a Lorentz invariant scalar 
\begin{equation}
\phi'=\omega' t'\!-\!k'x'=\omega t\!-\!kx=\phi
\tag{01}
\end{equation}
propagating with speed $\:\mathrm{w'}=\omega'/k'\:$ relatively to a frame $\:\mathrm{S'}$. If $\:\mathrm{S'}$ is moving with speed $\:-\upsilon\:$ relatively to $\:\mathrm{S}\:$ then
\begin{equation}
\left(\dfrac{c^2}{\mathrm{w'}}\right)=\dfrac{\upsilon+\left(\dfrac{c^2}{\mathrm{w}}\right)}{1+\dfrac{\upsilon}{\mathrm{w}}}=\dfrac{\upsilon+\left(\dfrac{c^2}{\mathrm{w}}\right)}{1+\dfrac{\upsilon\left(\dfrac{c^2}{\mathrm{w}}\right)}{c^2}}
\tag{02}
\end{equation}
defining
\begin{equation}
u\stackrel{def}{\equiv}\left(\dfrac{c^2}{\mathrm{w}}\right)\,,\quad u'\stackrel{def}{\equiv}\left(\dfrac{c^2}{\mathrm{w'}}\right)
\tag{03}
\end{equation}
equation (02) remind us the relativistic sum of particle velocities $\:\upsilon, u$ 
\begin{equation}
u'=\dfrac{\upsilon+u}{1+\dfrac{\upsilon u}{c^2}}
\tag{04}
\end{equation}
On the other hand this same equation (02) expressed as 
\begin{equation}
\quad \mathrm{w'}=\dfrac{\upsilon+\mathrm{w}}{1+\dfrac{\upsilon \mathrm{w}}{c^2}}
\tag{02'}
\end{equation}
remind us the relativistic sum of particle velocities but now this equation is valid for superluminal speeds also ($\:\mathrm{w}>c,\mathrm{w'}>c\:$). 
Now, if the phase $\:\phi=\omega t\!-\!kx\:$ is "superluminal" : $\:\mathrm{w}=\omega/k>c\:$, then it has this property in all frames(1) and the velocities $\:u, u'\:$ in (03) are "subluminal":  $\:u<c, u'<c\:$
corresponding to a particle and vice-versa : to a particle with speed $\:u<c\:$ there corresponds an accompanied "superluminal" plane phase wave propagating with speed $\:\mathrm{w}=c^2/u>c$. And this picture is Lorentz invariant, that is the same in all frames. Also, the product of the particle speed by the wave speed is invariant 
\begin{equation}
u'\cdot w'=c^2=u\cdot w
\tag{04'}
\end{equation}

(1)
The Lorentz transformation between the two aforementioned systems $\:\mathrm{S},\mathrm{S'}$ for the space-time coordinates $\:\left(x,t\right)\:$ is
\begin{align}
x &=\gamma_{\upsilon}\left(x'-\upsilon t'\right)
\tag{05.1}\\
t &=\gamma_{\upsilon}\left(t'-\dfrac{\upsilon x'}{c^2}\right)
\tag{05.2}
\end{align}
and by inversion ($\upsilon \rightarrow -\upsilon$)
\begin{align}
x' &=\gamma_{\upsilon}\left(x+\dfrac{\upsilon}{c}c t\right)
\tag{06.1}\\
ct' &=\gamma_{\upsilon}\left(ct+\dfrac{\upsilon}{c}x\right)
\tag{06.2}
\end{align}
For the plane phase wave
\begin{align} 
\phi \left(x,t\right) & = \omega \; t -k\; x= 
 \omega \underbrace{\left[\gamma_{\upsilon}\left(t'-\dfrac{\upsilon x'}{c^2}\right)\right]}_{t} -k\,\underbrace{\biggl[\gamma_{\upsilon}\left(x'-\upsilon t'\right)\biggr]}_{x}
\nonumber\\
& = \underbrace{\biggl[ \gamma_{\upsilon}\left(\omega+k\,\upsilon\right)\biggr]}_{\omega'} t^{\boldsymbol{\prime}} - \underbrace{\biggl[ \gamma_{\upsilon}\left(k+\dfrac{\upsilon \omega}{c^2}\right)\biggr]}_{k'} x' = \omega'\, t' - k'\,x'=\phi'\left( x',t'\right)
\tag{07}
\end{align}
that is
\begin{align}
ck' &=\gamma_{\upsilon}\left(ck+\dfrac{\upsilon}{c}\omega\right)
\tag{08.1}\\
\omega' &=\gamma_{\upsilon}\left(\omega+\dfrac{\upsilon }{c}ck\right)
\tag{08.2}
\end{align}
From (08) we see that the 2-vector
\begin{equation}
\boldsymbol{\Omega} \stackrel{def}{\equiv} \left(\omega,ck \right) 
\tag{09}  
\end{equation}
is transformed as the space-time 2-vector 
\begin{equation}
\mathbf{X} = \left(ct,x\right) 
\tag{10}  
\end{equation}
The phase is a Lorentz invariant as the inner product  of two vectors in Minkowski space
\begin{equation}
\phi '\left( x',t'\right)= \omega' t' - k'x'
= \dfrac{1}{c} \left(\boldsymbol{\Omega}^{\boldsymbol{\prime}}\boldsymbol{\cdot}\mathbf{X}^{\boldsymbol{\prime}} \right)= \dfrac{1}{c} \left(\boldsymbol{\Omega}\boldsymbol{\cdot}\mathbf{X} \right)= \omega \; t -k x=\phi \left(x,t\right) 
\tag{11}  
\end{equation}
Dividing equations (08) we have
\begin{equation}
\left(\dfrac{c^2 k'}{\omega'}\right)=\dfrac{\left(\dfrac{c^2 k}{\omega}\right)+\upsilon}{1+\dfrac{\upsilon \left(\dfrac{c^2 k}{\omega}\right)}{c^2}}
\tag{12}
\end{equation}
Having in mind that the speed of the phase wave is $\:\mathrm{w}=\omega/k \:$ in frame $\:\mathrm{S}\:$ and $\:\mathrm{w'}=\omega'/k'\:$ in frame $\:\mathrm{S'}\:$ your equation (02) is proved
\begin{equation}
\left(\dfrac{c^2}{\mathrm{w'}}\right)=\dfrac{\upsilon+\left(\dfrac{c^2}{\mathrm{w}}\right)}{1+\dfrac{\upsilon}{\mathrm{w}}}=\dfrac{\upsilon+\left(\dfrac{c^2}{\mathrm{w}}\right)}{1+\dfrac{\upsilon\left(\dfrac{c^2}{\mathrm{w}}\right)}{c^2}}
\tag{02}
\end{equation}



(2)
In the Figure 02 below all the points of a rectangle are turned ON simultaneously at time moment $\:t_{1}\:$ in frame $\:\mathrm{S}\:$ and remain on this ON state for $\:t\ge t_{1}$. We could say that this is a plane phase wave with  $\:\mathrm{w}=\infty\:$, wave length $\:\lambda=\infty\:$, that is $\:k=2\pi/\lambda=0\:$.


In frame $\:\mathrm{S'}\:$ the points of the rectangle are turned ON gradually by the front of a "superluminal" plane phase wave propagating with speed $\:\mathrm{w'}=c^2/\upsilon >c$. The turning ON is completed in a time interval $\:\Delta t'=\gamma_{\upsilon}\upsilon \ell/c^2\:$ where $\:\ell\:$ the width of the rectangle in frame $\:\mathrm{S}\:$ parallel to the velocity $\:\boldsymbol{\upsilon}$.


A: In addition to @robphy's answer, I kept searching for some wave phenomenon where the composition law for inverse velocity in my question plays a role, and I think I found one.
Consider the interference of two light waves propagating in opposite directions and with different frequencies,
$$\begin{align}
&\cos \omega_1\left(t - \frac{x}{c}\right) + \cos \omega_2\left(t + \frac{x}{c}\right)\\
=& 2\cos\left(\omega t - \frac{\Delta\omega}{c}x\right)\cos\left(\Delta\omega t - \frac{\omega}{c}x\right)
\end{align}$$
where
$$\begin{align}
\omega &= \frac{\omega_1+\omega_2}{2},\\
\Delta\omega&=\frac{\omega_1-\omega_2}{2}.
\end{align}$$
The phase velocities for respectively the phase of the first and the second cosine are
$$\begin{align}
v_1 &= c\frac{\omega}{\Delta\omega},\\
v_2 &= c\frac{\Delta\omega}{\omega}.
\end{align}$$
Being phase velocities, both of them transform according to the relativistic velocity addition formula, but then 
$$v_1 v_2 = c^2.$$
