# 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.

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