The value of $\dfrac{\sin ^2 20^{\circ}+\cos ^4 20^{\circ}}{\sin ^4 20^{\circ}+\cos ^2 20^{\circ}}$ is

Let $\theta = 20^{\circ}$. The expression is $E = \dfrac{\sin^{2}\theta + \cos^{4}\theta}{\sin^{4}\theta + \cos^{2}\theta}$.

Using the identity $\sin^{2}\theta = 1 - \cos^{2}\theta$, we substitute into the numerator:

Numerator $= (1 - \cos^{2}\theta) + \cos^{4}\theta = 1 - \cos^{2}\theta + \cos^{4}\theta$.

Using the identity $\cos^{2}\theta = 1 - \sin^{2}\theta$, we substitute into the denominator:

Denominator $= \sin^{4}\theta + (1 - \sin^{2}\theta) = 1 - \sin^{2}\theta + \sin^{4}\theta$.

Since $\cos^{2}\theta = 1 - \sin^{2}\theta$, the numerator becomes $1 - (1 - \sin^{2}\theta) + (1 - \sin^{2}\theta)^{2} = 1 - 1 + \sin^{2}\theta + 1 - 2\sin^{2}\theta + \sin^{4}\theta = 1 - \sin^{2}\theta + \sin^{4}\theta$.

Thus, the numerator is equal to the denominator, so $E = 1$.

Answer: $1$