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For $\alpha\in(0^\circ;90^\circ)$ simplify $E=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\sin\alpha+\cos\alpha}{\tan^2\alpha-1}.$
My try: $E=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\sin\alpha+\cos\alpha}{\dfrac{\sin^2\alpha-\cos^2\alpha}{\cos^2\alpha}}=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\sin\alpha\cdot\cos^2\alpha+\cos^3\alpha}{\sin^2\alpha-\cos^2\alpha}.$ Is there a better approach? Thank you in advance!
$$\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\sin\alpha+\cos\alpha}{\tan^2\alpha-1}=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\cos^2\alpha(\sin\alpha+\cos\alpha)}{\sin^2\alpha-\cos^2\alpha}=$$ $$=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\cos^2\alpha}{\sin\alpha-\cos\alpha}=\sin\alpha+\cos\alpha.$$
