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((sin^2y)/(1-cosy)) I have to rewrite this equation so it is not in a fraction form
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Hint'; 1/a = a^(-1)
Try the Pythagorean identity, which states that \[ \sin^2 y + \cos^2 y = 1. \]Using that, you get \[ \frac {\sin^2 y}{1 - \cos y} = \frac {1 - \cos^2 y}{1 - \cos y}. \]Now, recall difference of squares, which states that \[ \frac {1-a^2}{1 - a} = 1 + a. \]Can you finish from here?
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