<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] A variation is to apply the identity  $ \ \sin^4 x \ - \ \cos^4 \ = \ \sin^2 x \ - \ \cos^2 x \ $ , as Clive Newstead shows (through the Pythagorean Identity), to write
$$ \frac{2 \ (\sin^2 x \ - \cos^2 x) \ + \ \cos^2 x}{\sin^2 x \ + \ (2 \sin^2x - 1) } \ = \ \frac{2 \ (- \cos \ 2x) \ + \ \frac{1}{2} (1 + \cos \ 2x)}{\frac{1}{2} (1 - \cos \ 2x) \ + \ (- \cos \ 2x) } \ \ , $$
by applying the "double-angle" formulas for cosine and the  "sine-squared" and "cosine-squared" identities.
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