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#

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