$L^{-1}\left[ф\left(s\right)\right]\ =\ -\dfrac{1}{t}\ L^{-1}\left[ф^{'}\left(s\right)\right] \\[4ex] \begin{align*} \displaystyle \therefore{}L^{-1}{\ \tan }^{-1}\ \left(\dfrac{a}{s}\right)&=-\dfrac{1}{t}\ L^{-1}\left[\dfrac{d}{ds}{\tan }^{-1}\ \left(\dfrac{a}{s}\right)\right] \\[2ex] \displaystyle &=\ -\dfrac{1}{t}\ L^{-1}\left[\dfrac{1}{{1+\left(\dfrac{a}{s}\right)}^2}\ \left(\dfrac{-a}{s^2}\right)\right] \\[2ex] \displaystyle &=\ \ \dfrac{a}{t}\ L^{-1}\left[\ \dfrac{s^2}{{s^2+a}^2}\ \left(\dfrac{1}{s^2}\right)\right] \\[2ex] \displaystyle &=\ \ \dfrac{a}{t}\ {\ L}^{-1}\left(\ \dfrac{1}{{s^2+a}^2}\ \right) \\[2ex] \displaystyle &=\ \ \dfrac{a}{t}\ .\ \dfrac{1}{a}\sin{\left(at\right)} \\[2ex] \displaystyle &=\ \ \dfrac{\sin {\left(at\right)}}{t} \\[2ex] \end{align*}\\ \displaystyle {\ \therefore{}\ L}^{-1}\ {\tan }^{-1}\ \left(\dfrac{a}{s}\right)=\dfrac{\sin⁡(at)}{t} \\[2ex]$