If the force is given by $f = A\cos(Bx) + C\sin(Dt)$, find the dimensional formula of $AD/B$.

What to do in this kind of dimensional analysis problem

You use two facts: (1) force $f$ has dimensions $MLT^{-2}$, and (2) the inputs to $\sin(\cdot)$ and $\cos(\cdot)$ must be pure numbers (dimensionless).

Making the trigonometric arguments dimensionless

From $\cos(Bx)$, the argument $Bx$ must be dimensionless: $$[Bx]=1 \Rightarrow [B]=[x]^{-1}=L^{-1}.$$ From $\sin(Dt)$, the argument $Dt$ must be dimensionless: $$[Dt]=1 \Rightarrow [D]=[t]^{-1}=T^{-1}.$$

Using the fact that each term is a force

Since $f$ is a force and $\cos(Bx)$ is dimensionless, $$[A]=[f]=MLT^{-2}.$$ (Also $[C]=MLT^{-2}$, but it is not needed for $AD/B$.)

Dimensions of $AD/B$

Now combine the dimensions: $$\left[\frac{AD}{B}\right]=\frac{[A][D]}{[B]}=\frac{(MLT^{-2})(T^{-1})}{(L^{-1})}=ML^{2}T^{-3}.$$