# For $\alpha\in(0^\circ;90^\circ)$ simplify $E=\frac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\frac{\sin\alpha+\cos\alpha}{\tan^2\alpha-1}$

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!

• Correct.You can continue. Use the formula $a^2-b^2=(a-b)(a+b)$ Aug 29, 2020 at 15:30
• Thank you! I see it now. Aug 29, 2020 at 15:35
• You are welcome! Good luck :) Aug 29, 2020 at 15:37

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