Prove the following idennity: \( \frac{1-\cos ^{2} x+\sin (-2 x)}{\sin x-2 \cos x}=\sin x \)

Alright, let's break this down and prove the identity step by step! We start with the left-hand side: \[ \frac{1 - \cos^2 x + \sin(-2x)}{\sin x - 2 \cos x} \] Using the Pythagorean identity, we know that \(1 - \cos^2 x = \sin^2 x\). Also, recall that \(\sin(-2x) = -\sin(2x)\) due to the odd property of sine. So we can rewrite the expression: \[ \frac{\sin^2 x - \sin(2x)}{\sin x - 2 \cos x} \] Next, using the double angle identity, \(\sin(2x) = 2 \sin x \cos x\), we substitute that into our expression: \[ \frac{\sin^2 x - 2 \sin x \cos x}{\sin x - 2 \cos x} \] Now, factor the numerator: \[ \sin^2 x - 2 \sin x \cos x = \sin x(\sin x - 2 \cos x) \] So it becomes: \[ \frac{\sin x(\sin x - 2 \cos x)}{\sin x - 2 \cos x} \] As long as \(\sin x - 2 \cos x \neq 0\), we can simplify this to: \[ \sin x \] Thus, we have: \[ \frac{1 - \cos^2 x + \sin(-2x)}{\sin x - 2\cos x} = \sin x \] And there you have it! The identity is proven. ✨