<s>[INST] Proving $\frac{\sin 2x - \cos x}{4\sin^2x -1} = \frac{\sin^2x+\cos x+\cos^2x}{2\sin x +1} $ I need to prove that this is an identity:

$$\frac{\sin 2x - \cos x}{4\sin^2x -1} = \frac{\sin^2x+\cos x+\cos^2x}{2\sin x +1} $$

Here's what I'm confused on:


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*I noticed that the expression in the denominator for the first expression ($4sin^2(x) -1)$ looks like a double angle identity, but it also looks like a difference of squares. I know I can solve this as a difference of squares. Am I correct that this is a double angle identity, and if so, how can I solve it?

*I could rearrange the numerator for the second expression ($\sin^2x+\cos x+\cos^2x$) to sub in $1$ for $\sin^2\theta+\cos^2\theta$ (Pythagorean identity, where  $\sin^2\theta+\cos^2\theta=1$)
By doing so, I'd get $\cos x + 1$. But, this must be incorrect, for the final answer is $\dfrac{\cos x}{2\sin x-1} = \dfrac{\cos x}{2\sin x-1}$. That means the cosine must've been multiplied by $1$. How is that so?
(Here's the picture of the problem)

 [/INST] You cannot prove it because it is false:
$$\frac{\sin(2x) - \cos x}{4\sin^2(x) -1 }= \frac{2\sin x\cos x - \cos x}{(2\sin x -1)(2\sin x +1)} = \frac{\cos x}{2\sin x +1} \color{red}{\neq} \frac{1+\cos x}{2\sin x +1}$$
</s>