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.

Hint: Put the fractions over the same denominator $\left(\sin\left(\alpha\right)-\cos\left(\alpha\right)\right)\left(1-\tan^{2}\left(\alpha\right)\right)$, then eliminate the denominator on the left side. Now you have just to expand the left and right side, and finally use the identity $\tan\left(\alpha\right)=\frac{\sin\left(\alpha\right)}{\cos\left(\alpha\right)}.$ You will get the same expression.
