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Math Help - Can you show mi working for this

  1. #1
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    Question Can you show mi working for this

    (sec^4x - 2 sec^2x tan^2x + tan^4x) i need to simplify this equation and the answer is 1, i just dont know how 2 work it. Thx.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Sundae View Post
    (sec^4x - 2 sec^2x tan^2x + tan^4x) i need to simplify this equation and the answer is 1, i just dont know how 2 work it. Thx.
    Hi Sundae,

    I assume the expression you have is set to 0.

    \sec^4x-2\sec^2 x \tan^2 x+ \tan^4x=0

    Now factor:

    (\sec^2 x-\tan^2 x)^2=0

    Edit: No solution to the above equation, anyhow. So, you just want to simplify the expression. See my last post.
    Last edited by masters; April 27th 2009 at 09:27 AM.
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  3. #3
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    its not equal to 0, i just have to simplify till i get to 1.
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  4. #4
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    Quote Originally Posted by masters View Post
    Hi Sundae,

    I assume the expression you have is set to 0.

    \sec^4x-2\sec^2 x \tan^2 x+ \tan^4x=0

    Now factor:

    (\sec^2 x-\tan^2 x)^2=0
    Quote Originally Posted by Sundae View Post
    its not equal to 0, i just have to simplify till i get to 1.
    Ok sundae, now I understand.

    (\sec^2 x-\tan^2 x)^2

    (\sec^2 x-\tan^2 x)(\sec^2 x-\tan^2 x)

    \left(\frac{1}{\cos^2 x}-\frac{\sin^2 x}{\cos^2 x}\right)\left(\frac{1}{\cos^2 x}-\frac{\sin^2 x}{\cos^2 x}\right)

    \left(\frac{1-\sin^2 x}{\cos^2 x}\right)\left(\frac{1-\sin^2 x}{\cos^2 x}\right)


    \left(\frac{\cos ^2 x}{\cos^2 x}\right) \left(\frac{\cos ^2 x}{\cos^2 x}\right)=1
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  5. #5
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    Thumbs up

    Aw, that was really simple. Thankyou
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