Force F is given in terms of time t and distance x by $F = A \sin Ct + B \cos Dx$. Then the dimensions of $\dfrac {A}{B}$ and $\dfrac{C}{D}$ are:

$F = A \sin \, Ct + B \cos \, Dx$

$\therefore [Ct] = M^{0} L^{0} T^{0}$

$\Rightarrow[CT^1] = M^{0} L^{0} T^{0}$

$\Rightarrow [C] = T^{-1}$ $....(1)$

Similarly

$[Dx] = M^{0} L^{0} T^{0}$

$\Rightarrow$ $[DL^{1}] = M^{0} L^{0} T^{0}$

$\Rightarrow$ $[D] = L^{-1}$ $....(2)$

$[A] = [B] = MLT^{-2}$ $....(3)$

From equation $(3)$

$\dfrac{A}{B} = M^{0} L^{0} T^{0}$

From equation $(1)$ and $(2)$

$\dfrac{C}{D} = M^{0} L T^{-1}$

Hence Option $C$ is correct.