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    musicrulez's Avatar
    musicrulez Posts: 3, Reputation: 1
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    #1

    Jan 5, 2011, 11:43 PM
    Precalculus
    I have no idea how to solve this!
    It's about Verifying trigonometric identities.
    Here's the problem:

    1 divided by 1+tanx = cotx divided by 1+cotx
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    Jan 6, 2011, 05:47 AM

    Wouldn't it be easier to write:

    1/(1+tan(x))=cot(x)/(1+cot(x))

    Or, in LaTeX:



    To see the code I used to make it display that way, click on 'quote' at the lower right corner of this post.

    Just cross multiply:



    Now, can you see it? What does

    equal?
    musicrulez's Avatar
    musicrulez Posts: 3, Reputation: 1
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    #3

    Jan 11, 2011, 05:31 PM
    Comment on galactus's post
    THank you !
    musicrulez's Avatar
    musicrulez Posts: 3, Reputation: 1
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    #4

    Jan 11, 2011, 05:36 PM
    Comment on galactus's post
    So for another one, I have
    4tanxcos²x-2tanx/1-tan²x=2tanx/1-tan²x.
    So far, I have simplified it to 2tanxcos²x/1-tan²x.
    Could I set up a proportion setting my identities equal to each other?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Jan 11, 2011, 10:24 PM

    Are you allowed to cross multiply in proving identities :confused:

    My method would be to divide both sides by tan, or cot, depending on where you are starting from.






    This is using the fact that

    ~~~~~~~~~~~~~~~~


    I don't think this is a true identity, because:

    Multiply both sides by 1 - tan^2x gives a new identity:



    Adding 2 tan x to give this new identity:



    Factor 4tan x on the left:



    Simplifies to



    And as required for the identity to be true.

    Either that, or you did a typing mistake, the identity should be instead:


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