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.
