Approximation and Taylor Series Given that
$$ t^{+}=t^{-}+\Delta t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$
$$ c=\frac{L}{2}\Biggl(\frac{1}{t^{+}}+\frac{1}{t^{-}}\Biggl)\,\,\,\,\,\,\,\,(2) $$
how eqns (1) and (2) can be reworked to
$$  c=\frac{L}{t^{-}}\Biggl[\frac{1+\Delta t/t^{-}}{1+2\Delta t/t^{-}}\Biggl] \,\,\,\,\,\,\,\,(3)             $$
and why (3) can be reduced to
$$  c=\frac{L}{t^{-}}[1-\Delta t/t^{-}] \,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)             $$

I think there's a misplaced factor of two in your post, it should be
$$
c = \frac{L}{t^-} \left[ \frac{1 + \Delta t / 2t^-}{1 + \Delta t / t^-}\right]
$$
Now assume that $\Delta t / t^- \ll 1$, you can use the binomial approximation , keeping up to first order terms you get
\begin{eqnarray}
c &=& \frac{L}{t^-} \left(1 + \frac{\Delta t}{2t^-}\right) \color{blue}{\left(1 + \frac{\Delta t}{t^-}\right)^{-1}} \\
&\approx & \frac{L}{t^-} \left(1 + \frac{\Delta t}{2t^-}\right) \color{blue}{\left(1 - \frac{\Delta t}{t^-}\right)} \\
&=& \frac{L}{t^-}\left[1 - \frac{\Delta t}{t^-} + \frac{\Delta t}{2t^-} - \frac{1}{2}\left(\frac{\Delta t}{t^-}\right)^2 \right] \\
&\approx& \frac{L}{t^-}\left[1 - \frac{\Delta t}{2t^-} \right]
\end{eqnarray}
