Prove the identity: $$ \frac{-2 \sin ^{2} x+\cos x+1}{1-\cos \left(540^{\circ}-x\right)}=2 \cos x-1 $$

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We want to prove that

$$
\frac{-2\sin^2 x + \cos x + 1}{1-\cos(540^\circ-x)} = 2\cos x - 1.
$$

**Step 1. Rewrite the Denominator**

Notice that

$$
540^\circ = 360^\circ + 180^\circ.
$$

Thus,

$$
\cos(540^\circ-x)=\cos\big((360^\circ+180^\circ)-x\big)=\cos(180^\circ-x).
$$

It is known that

$$
\cos(180^\circ-x)=-\cos x.
$$

So, the denominator becomes

$$
1-\cos(540^\circ-x)=1-(-\cos x)=1+\cos x.
$$

**Step 2. Rewrite the Expression**

The given expression now is

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

Rearrange the numerator as

$$
1+\cos x - 2\sin^2 x.
$$

**Step 3. Use a Trigonometric Identity**

We can express $$\sin^2 x$$ in terms of $$\cos^2 x$$, using the Pythagorean identity:

$$
\sin^2 x = 1-\cos^2 x.
$$

Substitute into the numerator:

$$
1+\cos x-2(1-\cos^2 x)=1+\cos x-2+2\cos^2 x=2\cos^2 x+\cos x-1.
$$

**Step 4. Factor the Numerator**

Factor the quadratic expression $$2\cos^2 x + \cos x - 1$$. We look for factors of the form:

$$
(2\cos x-1)(\cos x+1).
$$

Check by expanding:

$$
(2\cos x-1)(\cos x+1) = 2\cos^2 x + 2\cos x - \cos x - 1 = 2\cos^2 x + \cos x - 1.
$$

So the numerator factors as

$$
(2\cos x-1)(\cos x+1).
$$

**Step 5. Simplify the Expression**

The expression is now:

$$
\frac{(2\cos x-1)(\cos x+1)}{1+\cos x}.
$$

Since the denominator $$1+\cos x$$ is the same as $$\cos x+1$$ (and assuming $$\cos x \neq -1$$ so that the cancellation is allowed), we cancel the common factor:

$$
\frac{(2\cos x-1)\,\cancel{(\cos x+1)}}{\cancel{(\cos x+1)}}=2\cos x-1.
$$

Thus, we have shown that

$$
\frac{-2\sin^2 x+\cos x+1}{1-\cos(540^\circ-x)}=2\cos x-1,
$$

which completes the proof.
$$ \frac{-2\sin^2 x + \cos x + 1}{1 - \cos(540^\circ - x)} = 2\cos x - 1 $$ is proven to be true.

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