Proving simple trigonometric identity I need help with this one
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha = \sin \alpha
$$
I tried moving sin a on the other side of the eqation
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha - \sin \alpha = 0
$$
This are the operations I was able to do
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\sin\alpha + \cos \alpha}{1- \frac{\sin^2\alpha}{\cos^2\alpha}} - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\sin\alpha + \cos \alpha}{\frac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha} } - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\sin\alpha*\cos^2\alpha + \cos^3 \alpha}{\cos^2\alpha- \sin^2\alpha} - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\cos^3 \alpha}{\sin\alpha} - \cos\alpha - \sin \alpha = 0
$$
I don't see what else I can do with this, so I tried to solve the left part of the equation.
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\sin\alpha + \cos \alpha}{1- \tan^2\alpha} - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\sin\alpha + \cos \alpha}{1- \frac{\sin^2\alpha}{\cos^2\alpha}} - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\sin\alpha + \cos \alpha}{\frac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha} } - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - cos\alpha}  + \frac{\sin\alpha\cdot\cos^2\alpha + \cos^3 \alpha}{\cos^2\alpha- \sin^2\alpha} - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha}  + \frac{\cos^3 \alpha}{\sin\alpha} - \cos\alpha = \sin \alpha
$$
And I get to nowhere again. I have no other ideas, I didn't see some formula or something. Any help is appreciated.

$\left(\dfrac{\sin^2 \alpha}{\sin \alpha+\cos\alpha}-(\sin\alpha+\cos\alpha)\right)+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\left(\dfrac{\sin^2 \alpha-\sin^2 \alpha+\cos^2\alpha}{\sin \alpha-\cos\alpha}\right)+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\dfrac{\cos^2\alpha}{\sin \alpha-\cos\alpha}+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\dfrac{\cos^2\alpha-\sin^2\alpha+\sin^2\alpha-\cos^2\alpha}{(\sin \alpha-\cos\alpha)(1-\tan^2\alpha)}\\=0$
