$\displaystyle Since\ f(t+2a)=f(t),\ f(t)\ is\ periodic\ with\ period\ function\ (A)=2a \\[2ex] $

$\displaystyle \therefore{}By\ definition\ Laplace\ transform\ for\ periodic\ function\ \\[2ex] $

$\displaystyle L\left[f(t)\right]=\frac{1}{1-e^{-As}}\int_0^Ae^{-st}f(t)dt \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{1-e^{-2as}}\left[\int_0^ae^{-st}\bullet{}1\ dt+\int_a^{2a}e^{-st}\bullet{}-1\ dt\right] \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{1-e^{-2as}}\left\{{\left[\frac{e^{-st}}{-s}\right]}_0^a-{\left[\frac{e^{-st}}{-s}\right]}_a^{2a}\right\} \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{1-e^{-2as}}\left\{\left[\frac{e^{-2as}}{-s}-\frac{1}{-s}\right]-\left[\frac{e^{-2as}}{-s}-\frac{e^{-sa}}{-s}\right]\right\} \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{1-e^{-2as}}\bullet{}\frac{1}{s}\left\{-e^{-as}+1+e^{-2as}-e^{-as}\right\} \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{s}\bullet{}\frac{\left[1-2e^{-as}+{\left(e^{-as}\right)}^2\right]}{1-{\left(e^{-as}\right)}^2} \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{s}\bullet{}\frac{{\left(1-e^{-as}\right)}^2}{\left(1-e^{-as}\right)\left(1+e^{-as}\right)} \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{s}\bullet{}\frac{{\left(1-e^{-as}\right)}^2}{\left(1+e^{-as}\right)} \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{s}\bullet{}\frac{e^{-\frac{as}{2}}}{e^{-\frac{as}{2}}}\frac{\left(e^{\frac{as}{2}}-e^{\frac{-as}{2}}\right)}{\left(e^{\frac{as}{2}}+e^{\frac{-as}{2}}\right)} \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{s}\bullet{}\frac{2\sinh{\left(\frac{as}{2}\right)}}{2\cosh{\left(\frac{as}{2}\right)}} \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{s}\bullet{}\tanh{\left(\frac{as}{2}\right)} \\[2ex]$