the argument, $\theta$ of cos or sin should be dimensionless.

therefore,

dimension of $\mathrm{Bx}=[M L T]$

$[B]\left[L^{\prime}\right]=[M L T]$

$[B]=\left[M L^{0} T\right]$

Similarly $[D]\left[T^{\prime}\right]=[M L T]$

$[D]=\left[M L T^{0}\right.$

dimension of $D/B=\frac{\left[M L T^{0}\right]}{\left[M L^{0} T\right]}$

$=\left[L^{1} T^{-1}\right]$