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Math Help - trig identities help !

  1. #1
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    trig identities help !

    hi everyone
    im ok with normal trig identiies but im confudes with theres any help please :

    heres the question :

    1) sin^4x - cos^4x = sin^2x - cos^2x

    2)sec^4x - tan^4x = 1 + 2tan^2x

    ...thanks
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  2. #2
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    Quote Originally Posted by mike--1988 View Post
    hi everyone
    im ok with normal trig identiies but im confudes with theres any help please :

    heres the question :

    1) sin^4x - cos^4x = sin^2x - cos^2x

    2)sec^4x - tan^4x = 1 + 2tan^2x

    ...thanks
    1) \sin^4{x} - \cos^4{x} = (\sin^2{x} + \cos^2{x})(\sin^2{x} - \cos^2{x})

     = 1(\sin^2{x} - \cos^2{x})

     = \sin^2{x} - \cos^2{x}.


    2) \sec^4{x} - \tan^4{x} = (\sec^2{x})^2 - \tan^4{x}

     = (1 + \tan^2{x})^2 - \tan^4{x}

     = 1 + 2\tan^2{x} + \tan^4{x} - \tan^4{x}

     = 1 + 2\tan^2{x}.
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  3. #3
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    Hello, mike--1988!


    Another approach to #2 . . .


    2)\;\sec^4\!x - \tan^4\!x \;=\; 1 + 2\tan^2\!x

    \text{Factor: }\:\sec^4\!x - \tan^4\!x \:=\:(\sec^2\!x + \tan^2\!x)\underbrace{(\sec^2\!x - \tan^2\!x)}_{\text{This is 1}}
    . . . . . . . . . . . . . . =\;\sec^2\!x + \tan^2\!x

    . . . . . . . . . . . . . . =\;(\tan^2\!x + 1) + \tan^2\!x

    . . . . . . . . . . . . . . = \;1 + 2\tan^2\!x

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