How to prove that kinetic energy equals $K=\frac{p^2}{(\gamma+1)m}$? In this video at timestamp 1:40:34, the lecturer states that the general form for kinetic energy expressed with momentum is
$$K = \frac{p^2}{ (\gamma + 1)m}$$
with $\gamma$ as the Lorentz factor.
which reduces in the non-relativistic case to the form we're all familiar with $K=(1/2)mv^2$ (since $p=\gamma m v$).
I've tried to prove this many times now, but for some reason I can't get to this result. Also I can't find this definition of kinetic energy anywhere else expressed like this. I can only find a way to express kinetic energy with momentum using the energy-momentum relation, which doesn't (I think) easily reduce to the form I stated here.
Here's something I tried:
\begin{align}
&K^2=(\gamma - 1)^2(mc^2)^2 \\
&=(\gamma - 1)^2(\gamma^2 m^2 c^4 - (pc)^2)\\
&= (\gamma - 1)^2c^2(\gamma^2 m^2 c^2 - p^2)\\
&\iff \\
&K = \frac{(\gamma - 1)^2 c^2(\gamma^2 m^2 c^2 - p^2)}{(\gamma - 1)c^2} \\
&= \frac{(\gamma - 1)(\gamma^2m^2c^2 - p^2)}{m}
\end{align}
which has some common terms, so I tried to expand and multiply (top & bottom) with $\gamma + 1$ but there are terms that do not vanish.
Can someone help me get the same result?
 A: Use rapidity $\theta$, so that $(v/c)=\tanh\theta$ and $\gamma=\displaystyle\frac{1}{\sqrt{1-(v/c)^2}}=\cosh\theta$.
These identities [derived from hyperbolic trigonometry] might help.
$$\gamma^2\left(1-(v/c)^2\right)=1 \quad \quad \mbox{from the definition of $\gamma$}$$
which can be solved for $(v/c)^2$ and written as
$$\mbox{as a difference of squares.... details omitted} $$
Start with the usual expression [in terms of $\gamma$] $$K=(\gamma-1)mc^2.$$
Can you take it from here?
(There is actually a more interesting way to write it.
I may update this post later.)

Here are some references for that expression you asked about.
D.E. Fahnline, "Parallels between relativistic and classical dynamics for introductory courses," Am.J.Phys. 43,  492--495 (1975).
https://doi.org/10.1119/1.9775
P.J.Riggs, "A Comparison of Kinetic Energy and Momentum in Special Relativity and Classical Mechanics,"
The.Phys.Teach. 54,  80--82 (2016).
https://doi.org/10.1119/1.4940169
T.A.Moore,
Six Ideas that Shaped Physics, Unit C, 2e.
(McGraw Hill, New York, 2003), Eq.(C8.16) on p.143.
R.Chabay and B.Sherwood,
Matter and Interactions, 4e
(Wiley, Hoboken NJ, 2015), p.220.

update:
Here's the underlying trigonometric identity (implied by the OP's problem):
$$(\cosh\theta-1)=\frac{\sinh^2\theta}{\cosh\theta+1},$$
which follows from
$$1\quad=\quad\cosh^2\theta-\sinh^2\theta \quad=\quad \cosh^2\theta(1-\tanh^2\theta).$$
A: One way that I found is to use the following two facts (along with $K = (\gamma-1)mc^2$):
$$E = \sqrt{p^2c^2 + m^2c^4}$$
$$E = \gamma mc^2$$
We thus have
$$\gamma mc^2 = \sqrt{p^2c^2 + m^2c^4} \implies \gamma^2m^2c^4 = p^2c^2 + m^2 c^4$$
Arranging this to get $p$ in terms of everything else,
$$(\gamma^2-1)m^2c^2 = p^2$$
Now, let's divide by $m$ on each side and note that $\gamma^2-1 = (\gamma+1)(\gamma-1)$ by simple expansion.
We then have
$$(\gamma+1)(\gamma-1)mc^2 = \frac{p^2}{m}$$
Using the fact that we have $K = (\gamma-1) mc^2$ it then follows that
$$K = \frac{p^2}{(\gamma + 1)m}$$
