We're trying to prove #sin^2x-sin^4x=cos^2x-cos^4x#.
Start with the left side and manipulate it until it looks like the right side. First, factor out the #sin^2x# term, then use the Pythagorean identity:
#sin^2x+cos^2x=1#
Here are some alternative (but equivalent) identities:
#sin^2x=1-cos^2x#
#cos^2x=1-sin^2x#
Now, here's the actual proof:
#LHS=color(purple)(sin^2x)-color(purple)(sin^4x)#
#color(white)(LHS)=color(blue)(sin^2x)*color(red)1-color(blue)(sin^2x)*color(red)(sin^2x)#
#color(white)(LHS)=color(blue)(sin^2x)(color(red)(1-sin^2x))#
#color(white)(LHS)=color(blue)((1-cos^2x))(color(red)(1-sin^2x))#
#color(white)(LHS)=color(blue)((1-cos^2x))(color(red)(cos^2x))#
#color(white)(LHS)=color(blue)((1-cos^2x))(color(red)(cos^2x))#
#color(white)(LHS)=color(blue)1*color(red)(cos^2x)-color(blue)(cos^2x)*color(red)(cos^2x)#
#color(white)(LHS)=color(purple)(cos^2x)-color(purple)(cos^4x)#
#color(white)(LHS)=RHS#