Correct Answer: $1$

Solution : Given:
$\sin^4\theta+\cos^4\theta=2\sin^2\theta \cos^2\theta$ and $\theta$ is an acute angle.
We know the algebraic identity, $(a-b)^2=a^2+b^2-2ab$.
$\sin^4\theta+\cos^4\theta-2\sin^2\theta \cos^2\theta=0$
Or, $(\sin^2\theta-\cos^2\theta)^2=0$
Or, $\sin^2\theta-\cos^2\theta=0$
Or, $\sin^2\theta=\cos^2\theta$
Or, $\tan^2\theta=1$
$\therefore \tan \theta=\pm 1$
Since, $\theta$ is an acute angle, then the value of $\tan\theta$ is $1$.
Hence, the correct answer is $1$.