Let force \(F=A \sin (C t)+B \cos (D x)\) where \(x\) and \(t\) are displacement and time respectively. The dimensions of \(\frac{\mathrm{C}}{\mathrm{D}}\) are same as dimensions of

\(\mathrm{F}=\mathrm{A} \sin \mathrm{Ct}+\mathrm{B} \cos \mathrm{Dx}\)
Since the argument of trigonometric functions is the angle, which is dimensionless
\(
\begin{aligned}
& \therefore[\mathrm{Ct}]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \\
& \Rightarrow\left[\mathrm{CT}^{1}\right]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \\
& \Rightarrow[\mathrm{C}]=\mathrm{T}^{-1}...(1)
\end{aligned}
\)
Similarly
\(
\begin{aligned}
&[\mathrm{Dx}]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \\
\Rightarrow &\left[\mathrm{DL}^{1}\right]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \\
\Rightarrow &[\mathrm{D}]=\mathrm{L}^{-1}...(2)
\end{aligned}
\)