$\begin{align*} \displaystyle L(\cos \ at\ \cos h\ at)\ &=\ L\left(\cos \ at\ \frac{e^{at}+\ e^{-at}\ }{2}\right) \\[2ex] \displaystyle &=\ \frac{1}{2}\ \left[L\left(e^{at}\cos \ at\ \right)\ +\ L\left(e^{-at}\cos \ at\ \right)\right] \\[2ex] \end{align*}$

$ \displaystyle L\left(\cos \ at\ \right)\ =\ \frac{s}{{s^2+a}^2} \\[4ex] By \ First \ shifting \ theorem,\\[4ex] \displaystyle L\left(e^{at}\cos \ at\ \right)=\ \frac{s-a}{{{\left(s-a\right)}^2+a}^2}\ =\ \frac{s-a}{{s^2-2as+2a}^2} \\[2ex] \displaystyle L\left(e^{-at}\cos \ at\ \right)=\ \frac{s+a}{{{\left(s+a\right)}^2+a}^2}\ =\ \frac{s+a}{{s^2+2as+2a}^2} \\[2ex] \displaystyle \therefore{}\ L(\cos \ at\ \cos h\ at)\ =\ \frac{1}{2}\ \left[L\left(e^{at}\cos \ at\ \right)\ +\ L\left(e^{-at}\cos \ at\ \right)\right] \\[2ex] \displaystyle =\ \frac{1}{2}\ \left[\frac{s-a}{\left({s^2+2a}^2\right)-2as}\ +\ \frac{s+a}{\left({s^2+2a}^2\right)+2as}\right] \\[2ex] \displaystyle =\ \frac{1}{2}\ \left[\frac{s\left(s^2+\ 2a^2+\ 2as\right)-a\left(s^2+\ 2a^2+\ 2as\right)+s\left(s^2+\ 2a^2-\ 2as\right)+a\left(s^2+\ 2a^2-\ 2as\right)}{{\left({s^2+2a}^2\right)}^2-4a^2s^2}\ \right] \\[2ex] \displaystyle =\ \frac{1}{2}\ \left[\frac{2s\left(s^2+\ 2a^2\right)-2a\left(\ 2as\right)}{s^4+4a^2s^2+4a^4-4a^2s^2}\ \right] \\[2ex] \displaystyle =\ \frac{s^3}{s^4+\ 4a^4} \\[2ex] \displaystyle \therefore{}\ L(\cos \ at\ \cos h\ at)\ =\ \frac{s^3}{s^4+\ 4a^4} \\[2ex] $