Consider a body of weight mg resting on a plane inclined at angle θ, with the applied force acting parallel to the surface. Resolving the weight into components along and perpendicular to the incline:

• Component tending to slide the body down the slope = mg·sin θ.
• Normal reaction, N = mg·cos θ.
• Maximum available (limiting) friction that resists motion = μN = μ·mg·cos θ, where μ is the coefficient of friction.

The body is on the point of slipping when the driving component just exceeds the friction the surface can supply:

mg·sin θ > μ·mg·cos θ ⇒ dividing both sides by mg·cos θ ⇒ tan θ > μ, i.e. μ < tan θ.

This is exactly the condition "coefficient of friction is less than the tangent of the inclination", so that option is correct. The special case μ = tan θ defines the angle of repose, at which the body is on the verge of sliding; below that (μ < tan θ) friction can no longer hold it and it slips.

The other statements do not describe the slip condition:

• Frictional force equals the weight of the body is not a meaningful slipping criterion; slipping depends on comparing the down-slope component with limiting friction, not on friction equalling the full weight.
• Normal reaction greater than weight cannot occur for a plain incline (N = mg·cos θ ≤ mg), and a larger N would actually raise the available friction, making slipping less likely, not more.
• Parallel force opposite to the frictional force merely describes a direction; on its own it says nothing about whether the driving effort exceeds limiting friction, which is what determines slipping.

Hence the body is most likely to slip when μ < tan θ.