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Math Help - Trig Simplification 2

  1. #1
    Member rtblue's Avatar
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    Trig Simplification 2

    Hey guys, I had another simplification question. You may start the process from either of these expressions, but the end result must the the other one:

    \displaystyle \frac{cscx}{secx-cscx}

    \displaystyle \frac{\cot^2x+cscxsecx+1}{\tan^2x-\cot^2x}

    Any suggestions would be appreciated! Thanks in advance!
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  2. #2
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    I would start by converting everything to sines and cosines...
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  3. #3
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    Hello, rtblue!

    \displaystyle \frac{\csc x}{\sec x-\csc x} \;=\;\frac{\cot^2\!x+\csc x\sec x+1}{\tan^2\!x-\cot^2\!x}

    I started with the right side:

    \displaystyle\frac{\cot^2\!x + \csc x\sec x + 1}{\tan^2\!x - \cot^2\!x} \;=\; \frac{\overbrace{\cot^2\!x + 1}^{\text{This is }\csc^2\!x} + \csc x\sec x}{(\sec^2\!x - 1) - (\csc^2\!x - 1)}

    . . \displaystyle =\;\frac{\csc^2\!x + \csc x \sec x}{\sec^2\!x - 1 - \csc^2\!x + 1} \;=\;\frac{\csc x\sec x + \csc^2\!x}{\sec^2\!x - \csc^2\!x}

    . . \displaystyle=\; \frac{\csc x(\sec x + \csc x)}{(\sec x - \csc x)(\sec x + \csc x)} \;=\;\frac{\csc x}{\sec x - \csc x}

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