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Math Help - Trig. Identities; Tricky one :S

  1. #1
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    Question Trig. Identities; Tricky one :S

    Prove:

    tanU/(secU+1) = (secU-1)/tanU

    i tried this many times, but i cant seem to prove them.
    I really need help with this, plz if can prove it, plz show your steps b/c I want to learn to do it myself for later questions.

    Thx
    Last edited by 2shoes; April 27th 2009 at 04:46 PM.
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  2. #2
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    LHS =  \frac{tanx}{secx+1} = \frac{\frac{sinx}{cosx}}{\frac{1}{cosx}+1} = \frac{sinx}{1+cosx} = LHS



    RHS = \frac{secx-1}{tanx} = \frac{\frac{1}{cosx}-1}{\frac{sinx}{cosx}} = \frac{\frac{1-cosx}{cosx}}{\frac{sinx}{cosx}}=

    = \frac{1-cosx}{sinx} * \color{red}\frac{(1+cosx)}{(1+cosx)} \color{black}= \frac{1-cos^2x}{sinx+sinxcosx} = \frac{sin^2x}{sinx(1+cosx)} = \frac{sinx}{1+cosx} = LHS

    The key for solving this and other similar identities is the red part. Hope that helps!
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  3. #3
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    Question woah...what just happned?

    LHS =  \frac{tanx}{secx+1} = \frac{\frac{sinx}{cosx}}{\frac{1}{cosx}+1} = \frac{sinx}{1+cosx} = LHS



    RHS = \frac{secx-1}{tanx} = \frac{\frac{1}{cosx}-1}{\frac{sinx}{cosx}} = \frac{\frac{1-cosx}{cosx}}{\frac{sinx}{cosx}}=

    Quote Originally Posted by Referos View Post
    = \frac{1-cosx}{sinx} * \color{red}\frac{(1+cosx)}{(1+cosx)} \color{black}= \frac{1-cos^2x}{sinx+sinxcosx} = \frac{sin^2x}{sinx(1+cosx)} = \frac{sinx}{1+cosx} = LHS

    The key for solving this and other similar identities is the red part. Hope that helps!
    why would u just times it by 1+cosx? and would that not change the whole thing, and make it a different answer? so they will not equal each other.
    could you plz expain why you did that.
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  4. #4
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    Quote Originally Posted by 2shoes View Post
    LHS =  \frac{tanx}{secx+1} = \frac{\frac{sinx}{cosx}}{\frac{1}{cosx}+1} = \frac{sinx}{1+cosx} = LHS



    RHS = \frac{secx-1}{tanx} = \frac{\frac{1}{cosx}-1}{\frac{sinx}{cosx}} = \frac{\frac{1-cosx}{cosx}}{\frac{sinx}{cosx}}=



    why would u just times it by 1+cosx? and would that not change the whole thing, and make it a different answer? so they will not equal each other.
    could you plz expain why you did that.
    No, because \frac{1 + \cos{x}}{1 + \cos{x}} = 1, and multiplying anything by 1 leaves it unchanged.


    This is the trick that's used in mathematics all the time - multiplying by a cleverly disguised 1 to change how something looks.
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  5. #5
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    Smile Thx

    No, because \frac{1 + \cos{x}}{1 + \cos{x}} = 1, and multiplying anything by 1 leaves it unchanged.


    This is the trick that's used in mathematics all the time - multiplying by a cleverly disguised 1 to change how something looks.
    haha thx, nice trick, never saw it before...again thx a lot
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  6. #6
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    Hello, 2shoes!

    Prove: . \frac{\tan x}{\sec x+1} \:=\:\frac{\sec x-1}{\tan x}

    Multiply by \frac{\sec x - 1}{\sec x - 1}

    . . \frac{\tan x}{\sec x + 1}\cdot\frac{\sec x - 1}{\sec x - 1} \;=\;\frac{\tan x(\sec x - 1)}{\underbrace{\sec^2\!x - 1}_{\text{This is }\tan^2\!x}} \;=\;\frac{\tan x(\sec x-1)}{\tan^2\!x} \;=\;\frac{\sec x - 1}{\tan x}

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