$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
$\mathcal{L} (\cos 2t) = \dfrac{s}{s^2 + 2^2}$
$\mathcal{L} (\cos 2t) = \dfrac{s}{s^2 + 4}$
$\mathcal{L} (t \cos 2t) = (-1)^1 \dfrac{d}{ds} \left[ \dfrac{s}{s^2 + 4} \right]$
$\mathcal{L} (t \cos 2t) = -\left[ \dfrac{(s^2 + 4)(1) - s(2s)}{(s^2 + 4)^2} \right]$
$\mathcal{L} (t \cos 2t) = -\dfrac{s^2 + 4 - 2s^2}{(s^2 + 4)^2}$
$\mathcal{L} (t \cos 2t) = -\dfrac{4 - s^2}{(s^2 + 4)^2}$
$\mathcal{L} (t \cos 2t) = \dfrac{s^2 - 4}{(s^2 + 4)^2}$ answer
