<s>[INST] Prove this identity: $\frac{2\sin^4x+\cos^2x-2\cos^4x}{3\sin^2x-1} =1$ I am stuck on this identity
$$\frac{2\sin^4x+\cos^2x-2\cos^4x}{3\sin^2x-1} =1$$
I began working on the left side trying to get things to cancel out or equal one by the Pythagorean identities. I am stuck and can't get it to reduce anymore. 
 [/INST] You want to prove
$$\frac{2\sin^4x+\cos^2x-2\cos^4x}{3\sin^2x-1} =1$$
which is equivalent to
$$2\sin^4x+\cos^2x-2\cos^4x=3\sin^2x-1$$
which is in turn equivalent to
$$(2\sin^4x+\cos^2x-2\cos^4x)-(3\sin^2x-1)=0.$$
Consider the function $f$ with $f(x)=(2\sin^4x+\cos^2x-2\cos^4x)-(3\sin^2x-1)$. We can compute the derivative:
$$
f'(x)=8\sin^3 x\,\cos x-2\cos x\sin x+8\cos^3x\sin x-6\sin x\cos x
$$
and we can pull out the common factor $8\sin x\cos x$ to rewrite this as
$$f(x)=-8\sin x\cos x(\sin^2x+\cos^2x-1)$$
which is clearly zero, so that $f$ is constant. As $f(0)=0$, as one can check at once, then $f(x)=0$ for all $x$, which proves what we wanted.
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