A block is placed on an inclined plane of inclination θ. The angle of inclination is such that the block slides down the plane at a constant speed. The coefficient of kinetic friction between the block and the inclined plane is equal to
tan θ
As the block moves down the inclined plane at a constant velocity, its acceleration will be zero. Hence, the net force acting on it will be zero.
mg sinθ=μN⇒mg sinθ=μmg cos θ⇒μ=tan θ
Hence, the correct choice is (c).