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Math Help - Identities

  1. #1
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    Identities

    Prove that identity tan2x - tanx = tanx sec2x


    I managed to start i think but am not sure

    (2tan x / 1-tan^2 x) - tan x

    but i don't know what else
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by gracey View Post
    Prove that identity tan2x - tanx = tanx sec2x


    I managed to start i think but am not sure

    (2tan x / 1-tan^2 x) - tan x

    but i don't know what else
    you may want to start with..

    \tan x \sec 2x = \frac{\tan x}{\cos 2x} = \frac{\tan x}{2\cos^2 x - 1} = \frac{\tan x}{\frac{2}{\sec^2 x}-1}

    = \frac{\tan x}{\frac{2 - \sec^2 x}{\sec^2 x}} = \frac{\tan x \sec^2 x}{2-\sec^2 x}

    this seems manageable..

    EDIT: from what you got,

    \frac{2\tan x}{1-\tan^2 x} - \tan x = \frac{2\tan x - \tan x + \tan^3 x}{1-\tan^2 x}
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  3. #3
    Senior Member nikhil's Avatar
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    Here it is

    Quote Originally Posted by gracey View Post
    Prove that identity tan2x - tanx = tanx sec2x


    I managed to start i think but am not sure

    (2tan x / 1-tan^2 x) - tan x

    but i don't know what else
    hi!gracey (sweet name)
    your starting is absolutly fine.
    (2tan x / 1-tan^2 x) - tan x
    now what you should really do is take tanx common
    tanx[(2/(1-tan^2x)-1]
    now use
    tan^2x=(1-cos2x)/1+cos2x
    after substituting we get
    tanx[2/(1-{(1-cos2x)/1+cos2x})-1]
    =tanx[{(1+cos2x)/cos2x}-1]
    =tanx[1/cos2x]
    =tanxsec2x
    hence proved
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  4. #4
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    Hello, gracey!

    A variation of nikhil's solution . . .


    Prove the identity: . \tan2x - \tan x \:=\: \tan x\sec2x

    I managed to start i think, but am not sure

    . . \frac{2\tan x}{1-\tan^2\!x} - \tan x

    Factor: . \tan x\left(\frac{2}{1-\tan^2\!x}-1\right) \;=\;\tan x\left(\frac{1+\tan^2\!x}{1-\tan^2\!x}\right)

    . . . . = \;\tan x\left(\frac{\sec^2\!x}{1-\tan^2\!x}\right) \;=\;\tan x\left(\frac{\frac{1}{\cos^2\!x}}{1 - \frac{\sin^2\!x}{\cos^2\!x}}\right)

    . . . . = \;\tan x\left(\frac{1}{\cos^2\!x - \sin^2\!x}\right) \;=\;\tan x\left(\frac{1}{\cos2x}\right)

    . . . . = \;\tan x\sec2x

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