= Solution
For a standard collinear <Lorentz boost>, put $\beta=v/c$ and $\Gamma=(1-\beta^2)^{-1/2}$. Its equations are $x'=\Gamma(x-\beta ct)$ and $ct'=\Gamma(ct-\beta x)$. Define the <rapidity> $\phi=\operatorname{arctanh}\beta$. Since $\cosh^2\phi-\sinh^2\phi=1$ and $\cosh\phi>0$, $\cosh\phi=\Gamma$ and $\sinh\phi=\Gamma\beta$. Therefore
$$
\boxed{x'=x\cosh\phi-ct\sinh\phi,\qquad
ct'=-x\sinh\phi+ct\cosh\phi}.
$$
Adding and subtracting, and using $\cosh\phi\mp\sinh\phi=e^{\mp\phi}$, gives
$$
\boxed{x'+ct'=e^{-\phi}(x+ct),\qquad x'-ct'=e^\phi(x-ct)}.
$$
A second boost multiplies these factors by $e^{\mp\phi'}$, so the combined <rapidity> is $\phi+\phi'$. The <hyperbolic tangent> addition formula then gives the <relativistic velocity addition> law
$$
\boxed{v''=c\tanh(\phi+\phi')=\frac{v+v'}{1+vv'/c^2}}.
$$
Velocities are signed along the common axis, so this formula also covers oppositely directed boosts.
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