Step 1 — Set up the right triangle.

$$\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{12}{13}$$

So $B = 12k$ and $H = 13k$.

Step 2 — Find the perpendicular:

$$P^2 = H^2 - B^2 = (13k)^2 - (12k)^2$$

$$= 169k^2 - 144k^2 = 25k^2$$

$$P = 5k$$

Step 3 — Find $\cot\theta$:

$$\cot\theta = \frac{B}{P} = \frac{12k}{5k} = \frac{12}{5}$$

$$\boxed{\cot\theta = \frac{12}{5}}$$

Check with the identity:

$$\operatorname{cosec}^2\theta - \cot^2\theta = \frac{169}{25} - \frac{144}{25} = 1 \quad \checkmark$$

Careful: $\cot\theta = \dfrac{\text{Base}}{\text{Perpendicular}}$. Getting $\frac{5}{12}$ (option c) means the ratio was inverted — that is $\tan\theta$.

(5, 12, 13 is a standard Pythagorean triplet.)

Correct option: (d) $\dfrac{12}{5}$