The original statement, with #theta# added for sensibility.
#(2cos^2theta-1)^2/(cos^4theta-sin^4theta)=1-2sin^2theta#
Let's work the left hand side.
Using one form of the cosine double angle formula, #cos(2theta)=2cos^2theta-1#, the left side becomes
#(cos(2theta))^2/(cos^4theta-sin^4theta)#, or equivalently #cos^2(2theta)/(cos^4theta-sin^4theta)#
Then, notice that the denominator is a difference of squares, it can be factored as follows: #a^2-b^2=(a+b)(a-b)#. In this case #a=cos^2theta# and #b=sin^2theta#. So the left side is now
#cos^2(2theta)/((cos^2theta+sin^2theta)(cos^2theta-sin^2theta))#
Now we can use the Pythagorean Identity, #sin^2theta+cos^2theta=1# to get
#cos^2(2theta)/((1)(cos^2theta-sin^2theta))#, or equivalently #cos^2(2theta)/(cos^2theta-sin^2theta)#
And if we can use another form of the cosine double angle formula, #cos(2theta)=cos^2theta-sin^2theta#, we can see that
#cos^2(2theta)/cos(2theta)#
Simplifying this gives a friendly
#cos(2theta)#
And using the third and final form of the cosine double angle formula, #cos(2theta)=1-2sin^2theta#, the left hand side finally becomes
#1-2sin^2theta#
And since #1-2sin^2theta=1-2sin^2theta#, the left side is equal to the right side and the identity is proven. Q.E.D.
