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Math Help - Help with a trig identity

  1. #1
    Junior Member
    Joined
    Nov 2007
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    Help with a trig identity

    This identity makes no sense to me:

    cot(x) - tan(x) - tan(x) + cot(x) = (4 - 2sec^2(x))/tan(x)

    I did this:

    LS
    =2cot(x) - 2tan(x)
    =2{cot(x) - tan(x)}
    =2[{1/tan(x)} - tan(x)]
    =2[{1-tan^2(x)}/tan(x)]
    =2{sec^2(x)}/tan(x)

    Where did i go wrong? I don't have a "4-" term.
    Please help...

    ps. I used the different bracket types to differentiate from different sets of brackets. i'm sorry if it means something else in math...
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, CenturionMonkey!

    \cot x - \tan x - \tan x + \cot x \:= \:\frac{4 - 2\sec^2\!x}{\tan x}


    I did this:

    \text{LS}\:=\:2\cot x  - 2\tan x \:=\:2(\cot x - \tan x) \;=\;2\left(\frac{1}{\tan x} - \tan x \right)

    . . =\:2\left(\frac{1-\tan^2\!x}{\tan x}\right) \:=\:\frac{2\,{\color{red}\sec^2\!x}}{\tan x} . . . . no

    We have: . 2\left(\frac{1 - \tan^2\!x}{\tan x}\right)\;=\;2\left(\frac{1-(\sec^2\!x-1)}{\tan x}\right) \;=\;2\left(\frac{2-\sec^2\!x}{\tan x}\right)

    Therefore: . \frac{4-2\sec^2\!x}{\tan x}

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  3. #3
    Junior Member
    Joined
    Nov 2007
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    Oh right! cause 1 + tan^2(x) = sec^2(x).

    thanks a lot
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