Let the ladder touch the wall at the top and the floor at the bottom.

• At the wall: normal reaction $N_1$ is horizontal.
• At the floor: normal reaction $N_2$ is vertical.
• Since the ladder is about to slip, friction at both contacts (if present) will be limiting.

We analyze all possible cases.

1. Forces on the ladder

Take the ladder to tend to slip such that the bottom moves away from the wall and the top moves downward. Then:

• Friction at the wall acts upward: $f_1$
• Friction at the floor acts towards the wall: $f_2$

At limiting equilibrium, $f_1 \le \mu_1 N_1, \qquad f_2 \le \mu_2 N_2$ and if friction is active at the limiting value, $f_1 = \mu_1 N_1, \qquad f_2 = \mu_2 N_2.$

2. Force balance

Horizontal equilibrium

$N_1 = f_2$ If floor friction is limiting, N_1 = \mu_2 N_2 \tag{1}

Vertical equilibrium

$N_2 + f_1 = mg$ If wall friction is limiting, N_2 + \mu_1 N_1 = mg \tag{2}

Substitute $N_1 = \mu_2 N_2$ into (2): $N_2 + \mu_1 \mu_2 N_2 = mg$ $N_2(1+\mu_1\mu_2)=mg$ $\boxed{N_2 = \frac{mg}{1+\mu_1\mu_2}}$

This matches Option C.

3. Torque balance about the bottom end

Taking moments about the bottom end:

• Weight $mg$ acts at the midpoint, giving anticlockwise moment $mg \cdot \frac{L}{2} \cos\theta$
• $N_1$ at the top gives clockwise moment $N_1 \cdot L\sin\theta$
• Friction at wall $f_1$ at the top gives anticlockwise moment $f_1 \cdot L\cos\theta$

So, $N_1L\sin\theta = mg\frac{L}{2}\cos\theta + f_1L\cos\theta$ $N_1\sin\theta = \left(\frac{mg}{2}+f_1\right)\cos\theta$ N_1\tan\theta = \frac{mg}{2}+f_1 \tag{3}

Now consider special cases.

4. Check Option D

Option D says:

• $\mu_1 = 0$
• $\mu_2 \ne 0$
• $N_1\tan\theta = \dfrac{mg}{2}$

If $\mu_1=0$, then wall is smooth, so $f_1=0$ Then from (3), $N_1\tan\theta = \frac{mg}{2}$ which is exactly Option D.

So D is correct.

5. Check Option A

Option A says:

• $\mu_1=0$
• $\mu_2\ne 0$
• $N_2\tan\theta = \dfrac{mg}{2}$

But with $\mu_1=0$, vertical equilibrium gives $N_2 = mg$ So Option A would imply $mg\tan\theta = \frac{mg}{2}$ which is not generally true.

Hence A is incorrect.

6. Check Option B

Option B says:

• $\mu_1\ne 0$
• $\mu_2=0$
• $N_1\tan\theta = \dfrac{mg}{2}$

If $\mu_2=0$, floor is smooth, so no horizontal friction at floor: $f_2=0$ Horizontal equilibrium gives $N_1=0$ But then the ladder cannot remain in general equilibrium against the wall unless trivial/impossible conditions occur. Also $N_1\tan\theta=0$, not $mg/2$.

Hence B is incorrect.

7. Final answer

Correct options are: $\boxed{\text{C, D}}$

This agrees with the stored answer.