Prove $\sin^2 \theta +\cos^4 \theta =\cos^2 \theta +\sin^4 \theta $

Rewrite this as

$$ \sin^2 \theta - \cos^2 \theta = \sin^4 \theta - \cos^4 \theta $$ and then factor the right-hand side as a difference of two squares.

As $\cos^2 \theta +\sin^2 \theta= 1$ we have $$\cos^4 \theta -\sin^4 \theta =(\cos^2 \theta -\sin^2 \theta)\color{red}{(\cos^2 \theta +\sin^2 \theta)}= \cos^2 \theta -\sin^2 \theta$$ That is,

$$\sin^2 \theta +\cos^4 \theta =\cos^2 \theta +\sin^4 \theta$$

I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $\sin^2\theta + \cos^2\theta = 1 $ rearranged into $\sin^2\theta = 1 - \cos^2\theta $ and $\cos^2\theta = 1 - \sin^2\theta $

We can see that: $\cos^4\theta = \cos^2\theta\cos^2\theta = (1-\sin^2\theta)(1-\sin^2\theta) = 1-2\sin^2\theta + \sin^4\theta $

$\sin^2\theta + \cos^4\theta = \cos^2\theta + \sin^4\theta $

$\sin^2\theta + (1-2\sin^2\theta + \sin^4\theta) = \cos^2\theta + \sin^4\theta $

$\sin^2\theta + 1-2\sin^2\theta + \sin^4\theta = \cos^2\theta + \sin^4\theta $

$\sin^4\theta-\sin^2\theta+1= \cos^2\theta + \sin^4\theta $

$\sin^4\theta-(1-\cos^2\theta)+1=\cos^2\theta + \sin^4\theta $

$\sin^4\theta-1+\cos^2\theta+1=\cos^2\theta + \sin^4\theta $

$\sin^4\theta+\cos^2\theta=\cos^2\theta + \sin^4\theta $