Prove that: (6) \( \frac{\cos 2 \alpha+2 \sin \alpha-1}{2 \cos \alpha-\sin 2 \alpha}=\tan \alpha \)

To prove that \[ \frac{\cos 2\alpha + 2\sin \alpha - 1}{2\cos \alpha - \sin 2\alpha} = \tan \alpha, \] we start with the identities for sine and cosine. We know: 1. \(\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = 2\cos^2 \alpha - 1\) (using the double angle formula). 2. \(\sin 2\alpha = 2\sin \alpha \cos \alpha\). Now substituting these into the left-hand side: The numerator becomes: \[ \cos 2\alpha + 2\sin \alpha - 1 = (2\cos^2 \alpha - 1) + 2\sin \alpha - 1 = 2\cos^2 \alpha + 2\sin \alpha - 2. \] The denominator becomes: \[ 2\cos \alpha - \sin 2\alpha = 2\cos \alpha - 2\sin \alpha \cos \alpha = 2\cos \alpha (1 - \sin \alpha). \] Putting it all together: \[ \frac{2\cos^2 \alpha + 2\sin \alpha - 2}{2\cos \alpha(1 - \sin \alpha)} = \frac{2(\cos^2 \alpha + \sin \alpha - 1)}{2\cos \alpha(1 - \sin \alpha)} = \frac{\cos^2 \alpha + \sin \alpha - 1}{\cos \alpha(1 - \sin \alpha)}. \] Now we simplify \(\cos^2 \alpha + \sin \alpha - 1\): Using \(\cos^2 \alpha = 1 - \sin^2 \alpha\): \[ \Rightarrow 1 - \sin^2 \alpha + \sin \alpha - 1 = -\sin^2 \alpha + \sin \alpha = -\sin \alpha(\sin \alpha - 1). \] Substituting back, we have: \[ \frac{-\sin \alpha(\sin \alpha - 1)}{\cos \alpha(1 - \sin \alpha)}. \] This can be rearranged to: \[ =-\frac{\sin \alpha}{\cos \alpha} \cdot \frac{\sin \alpha - 1}{1 - \sin \alpha}. \] Note that \(\frac{\sin \alpha - 1}{1 - \sin \alpha} = -1\), thus we get: \[ \frac{\sin \alpha}{\cos \alpha} = \tan \alpha. \] Hence, we conclude that \[ \frac{\cos 2\alpha + 2 \sin \alpha - 1}{2 \cos \alpha - \sin 2\alpha} = \tan \alpha. \] This proves the original statement.