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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!

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    $\begingroup$ Correct.You can continue. Use the formula $a^2-b^2=(a-b)(a+b)$ $\endgroup$ Aug 29, 2020 at 15:30
  • $\begingroup$ Thank you! I see it now. $\endgroup$
    – 10th grade
    Aug 29, 2020 at 15:35
  • $\begingroup$ You are welcome! Good luck :) $\endgroup$ Aug 29, 2020 at 15:37

1 Answer 1

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

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