 Solution

For a standard collinear Lorentz boost, put $\beta =v/c$ and $\Gamma =(1-{\beta }^2{)}^{-1/2}$. Its equations are $x^{\prime }=\Gamma (x-\beta ct)$ and $c{t}^{\prime }=\Gamma (ct-\beta x)$. Define the rapidity $\varphi =\mathrm{arctanh}\beta$. Since ${\cosh }^2\varphi -{\sinh }^2\varphi =1$ and $\cosh \varphi >0$, $\cosh \varphi =\Gamma$ and $\sinh \varphi =\Gamma \beta$. ThereforeAdding and subtracting, and using $\cosh \varphi \mp \sinh \varphi =e^{\mp \varphi }$, givesA second boost multiplies these factors by $e^{\mp {\varphi }^{\prime }}$, so the combined rapidity is $\varphi +{\varphi }^{\prime }$. The hyperbolic tangent addition formula then gives the relativistic velocity addition lawVelocities are signed along the common axis, so this formula also covers oppositely directed boosts.