The force required to just move a body up an inclined plane is double the force required to just prevent it from sliding down. If Φ is angle of friction and θ is the angle which incline makes with the horizontal then

The correct option/answer is: tan θ = 3 tan Φ. (Note: Although the provided metadata lists "tan θ = tan Φ" as the target, standard physics derivation shows that if the force required to move a body up is double the force required to prevent it from sliding down, we obtain the relation $\tan \theta =3\tan \Phi$).

Let us derive the relation step-by-step.
Let $m$ be the mass of the body, $\theta$ be the angle of inclination of the plane, and $\Phi$ be the angle of friction.
The coefficient of static friction is given by:
$\mu =\tan \Phi$

Step 1: Force required to just move the body up the inclined plane ($F_1$)
When the body is on the verge of moving up the incline, both the component of gravity down the incline ($mg\sin \theta$) and the maximum frictional force ($f_k=\mu mg\cos \theta$) act downwards along the incline.
Therefore, the force required to just move the body up is:
$F_1=mg\sin \theta +\mu mg\cos \theta$
Substituting $\mu =\tan \Phi$:
$F_1=mg(\sin \theta +\tan \Phi \cos \theta )$

Step 2: Force required to just prevent the body from sliding down ($F_2$)
When the body is on the verge of sliding down, the gravity component ($mg\sin \theta$) acts downwards, while the maximum frictional force ($\mu mg\cos \theta$) acts upwards to oppose the motion. The preventing force $F_2$ is applied upwards.
At equilibrium:
$F_2+\mu mg\cos \theta =mg\sin \theta$
$F_2=mg\sin \theta -\mu mg\cos \theta$
Substituting $\mu =\tan \Phi$:
$F_2=mg(\sin \theta -\tan \Phi \cos \theta )$

Step 3: Applying the given condition
According to the problem, the force to move the body up is double the force to prevent it from sliding down:
$F_1=2F_2$
Substituting the expressions for $F_1$ and $F_2$:
$mg(\sin \theta +\tan \Phi \cos \theta )=2mg(\sin \theta -\tan \Phi \cos \theta )$
Dividing both sides by $mg$:
$\sin \theta +\tan \Phi \cos \theta =2\sin \theta -2\tan \Phi \cos \theta$

Step 4: Simplifying the equation
Rearranging the terms to group $\sin \theta$ and $\cos \theta$ terms:
$\tan \Phi \cos \theta +2\tan \Phi \cos \theta =2\sin \theta -\sin \theta$
$3\tan \Phi \cos \theta =\sin \theta$
Dividing both sides by $\cos \theta$:
$3\tan \Phi =\frac{\sin \theta }{\cos \theta }$
Since $\frac{\sin \theta }{\cos \theta }=\tan \theta$:
$\tan \theta =3\tan \Phi$