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Math Help - Clarification

  1. #1
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    Clarification

    This is not strictly a calculus problem, but a question about the form of a trigonometric integral.

    Can someone tell me how -ln|csc x + cot x| is equal to ln|csc x - cot x|?

    I know that using the properties of logarithms that

    -ln|csc x + cot x| = ln (|csc x + cot x|)^-1 = ln |[1/(csc x + cot x)]|, but, I'm drawing a blank as to how any of these forms translates to
    ln|csc x - cot x|.
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  2. #2
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    Quote Originally Posted by kaiser0792 View Post
    This is not strictly a calculus problem, but a question about the form of a trigonometric integral.

    Can someone tell me how -ln|csc x + cot x| is equal to ln|csc x - cot x|?

    I know that using the properties of logarithms that

    -ln|csc x + cot x| = ln (|csc x + cot x|)^-1 = ln |[1/(csc x + cot x)]|, but, I'm drawing a blank as to how any of these forms translates to
    ln|csc x - cot x|.
    |\csc{x} + \cot{x}|^{-1} = \left|\frac{1}{\csc{x} + \cot{x}}\right|

     = \left|\frac{\csc{x} - \cot{x}}{(\csc{x} + \cot{x})(\csc{x} - \cot{x})}\right|

     = \left|\frac{\csc{x} - \cot{x}}{\csc^2{x} - \cot^2{x}}\right|


    But \csc^2{x} - \cot^2{x} = 1

    So \left|\frac{\csc{x} - \cot{x}}{\csc^2{x} - \cot^2{x}}\right| = \left|\frac{\csc{x} - \cot{x}}{1}\right|

     = |\csc{x} - \cot{x}|.
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  3. #3
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    Thank you Prove IT, I knew it was an algebra step, but I've been deep into trig integrals and couldn't see the forest for the trees!
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