Answer $A$
Given: $$\cos^2x + \cos^4x = 1$$
$$\cos^2x + \cos^4x = 1 \implies \cos^4x = 1 - \cos^2x = \sin^2x$$
$$\implies \cos^4x = \sin^2x\dots \qquad \to (1)$$
To find: $$\tan^2x + \tan^4x $$
$$\implies \tan^2x + \tan^4x = \frac{\sin^2x}{\cos^2x} + \frac{\sin^4x}{\cos^4x} \qquad \to (1)$$
$$\frac{\sin^2x}{\cos^2x} + \frac{\sin^4x}{\cos^4x} = \frac{\cos^4x}{\cos^2x} + \frac{\sin^4x}{\sin^2x} = \cos^2x + \sin^2x = 1$$
$\therefore A $ is the correct option.