On a banked road,

Resolving forces in horizontal and vertical directions, we have

$R \sin \theta+f \cos \theta=\frac{m v^2}{r}$

and $R \cos \theta-f \sin \theta =m g$ $\Rightarrow \quad \frac{R \sin \theta+f \cos \theta}{R \cos \theta-f \sin \theta}=\frac{v^2}{r g}$

$\Rightarrow \frac{\tan \theta+f / R}{1-f / R \tan \theta}=\frac{v^2}{r g}$

$\Rightarrow \frac{\tan \theta+\mu}{1-\mu \tan \theta}=\frac{v^2}{r g}$

$\Rightarrow \left(\mu v^2+r g\right) \tan \theta=v^2-\mu r g$

$\Rightarrow \tan \theta=\frac{v^2-\mu r g}{\mu v^2+r g}$