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. 
 A: Hint: Use the fact that
$$\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = \cdots$$
You'll see that things cancel nicely.
A: 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.
A: Let's take this one step at a time.
First, move the denominator to the right side.
1)  $$ 2sin^4(x) + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1 $$
Use the identity $ sin^2(x) = 1-cos^2(x) $ to rewrite $ sin^4(x) = (1-cos^2(x))^2$.
2)  $$ 2(1-cos^2(x))^2 + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1 $$
Now foil out $2(1-cos^2(x))^2 = 2(1 - 2cos^2(x) + cos^4(x))$ and distribute the 2.
3)  $$ 2-4cos^2(x) + 2cos^4(x) + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1$$
Now there are three things we can do here
First we have $ 2cos^4(x) $ and $ -2cos^4(x) $ so those will cancel out.  Second we have $ -4cos^2(x) + cos^2(x) = -3cos^2(x)$.  And last we can move the 2 from the left side of the equation to the right. 
4)  $$ -3cos^2(x) = 3sin^2(x) - 3 $$
Now factor out the 3 and it cancels out
5)  $$-cos^2(x) = sin^2(x) - 1 $$
And just rearrange it and we get
6)  $$ 1 = sin^2(x) + cos^2(x) $$
Hope that helps!
A: 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.
