The value of the expression 3 20^ - 4 20^ is equal to :

The given expression is $\sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ}$.

Converting cotangent to sine and cosine:
$= \dfrac{\sqrt{3} \cos 20^{\circ}}{\sin 20^{\circ}} - 4 \cos 20^{\circ}$

$= \dfrac{\sqrt{3} \cos 20^{\circ} - 4 \sin 20^{\circ} \cos 20^{\circ}}{\sin 20^{\circ}}$

Using the double angle formula $2 \sin \theta \cos \theta = \sin(2\theta)$ in the second term of the numerator:
$= \dfrac{\sqrt{3} \cos 20^{\circ} - 2 \sin 40^{\circ}}{\sin 20^{\circ}}$

Substitute $\sqrt{3} = 2 \sin 60^{\circ}$:
$= \dfrac{2 \sin 60^{\circ} \cos 20^{\circ} - 2 \sin 40^{\circ}}{\sin 20^{\circ}}$

Using the product-to-sum formula $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:
$2 \sin 60^{\circ} \cos 20^{\circ} = \sin(60^{\circ} + 20^{\circ}) + \sin(60^{\circ} - 20^{\circ}) = \sin 80^{\circ} + \sin 40^{\circ}$

Substituting this back into the numerator:
$= \dfrac{\sin 80^{\circ} + \sin 40^{\circ} - 2 \sin 40^{\circ}}{\sin 20^{\circ}}$

$= \dfrac{\sin 80^{\circ} - \sin 40^{\circ}}{\sin 20^{\circ}}$

Now, apply the sum-to-product formula $\sin C - \sin D = 2 \cos\left(\dfrac{C+D}{2}\right) \sin\left(\dfrac{C-D}{2}\right)$:
$\sin 80^{\circ} - \sin 40^{\circ} = 2 \cos 60^{\circ} \sin 20^{\circ}$

Since $\cos 60^{\circ} = \dfrac{1}{2}$:
$= 2 \left(\dfrac{1}{2}\right) \sin 20^{\circ} = \sin 20^{\circ}$

The expression becomes:
$= \dfrac{\sin 20^{\circ}}{\sin 20^{\circ}} = 1$

Answer: $1$