The coefficient of kinetic friction ${\mu }_s$ is given by
${\mu }_r=\frac{\text{ frictional force }}{\text{ normal reaction }}$ body of mass $M$ sliding down an inclined plane with $\theta$ as the angle of inclination has its coefficient of static friction as ${\mu }_s=\tan \theta$. This is proved below i.e.
(Free body diagram of the body for the body FBD)
Note: If the body slides down, the frictional force $(fr)$ act the opposition direction as shown so that
$\mathrm{\Sigma }{F}_{\text{net }}=ma-$ applying Newton's law
$mg\sin \theta -fr=ma$
$fr=mg\sin \theta -ma$ $fr=m(\sin \theta -a)$
$fr=m(g\sin \theta -a)$ but $fr=\mu mg\cos \theta$
but fr $=\mu mg\cos \theta$
$\mu mg\cos \theta =m(g\sin \theta -a),a=0$
$\mu mg\cos \theta =mg\sin \theta$
$\mu \cos \theta =\sin \theta$
$\mu s=\frac{\sin \theta }{\cos \theta }=\tan \theta$