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I got an urge for some trigonometry so I dug up my old books and got to it! I also have a lot of my old answers as well but there was one I can't figure out how I was thinking!

I had to simpify this expression:

$$\sin x + \frac{\cos^2 x}{\sin x} = \frac{\sin^2x+\cos^2x}{\sin x}=\frac{1}{\sin x}$$

I can't figure out I got from the first to second step.. Any suggestions? Thank you! :)

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    $\begingroup$ $\displaystyle a+{b\over c}={a\over 1}+{b\over c}={a\cdot c\over c}+{b\over c}={ac+b\over c} $. $\endgroup$
    – David Mitra
    Jun 7, 2015 at 10:15

1 Answer 1

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$$\sin x+\frac{\cos^2x}{\sin x}=\frac{\sin^2x}{\sin x}+\frac{\cos^2x}{\sin x}=\frac{\sin ^2x+\cos^2 x}{\sin x}=\frac{1}{\sin x}$$

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