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Math Help - please help me with hard trig identity

  1. #1
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    Exclamation please help me with hard trig identity

    Hi I have a quiz tomorrow and have been looking over some review problems that will be similar to those on the quiz. I've stumbled across a hard one and really need help. I've attempted numerous approaches and still haven't figured it out, so can someone walk me through?

    THANKS! (i removed x's for visual purposes)

    ((tan^2)-1)/(sec^2) = (tan-cot)/(tan+cot)
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TacticalPro View Post
    Hi I have a quiz tomorrow and have been looking over some review problems that will be similar to those on the quiz. I've stumbled across a hard one and really need help. I've attempted numerous approaches and still haven't figured it out, so can someone walk me through?

    THANKS! (i removed x's for visual purposes)

    ((tan^2)-1)/(sec^2) = (tan-cot)/(tan+cot)
    As usual, I have a way, but I doubt it's the most direct. Take a look at the RHS. Write everything in terms of sines and cosines, add the fractions in the numerator and denominator, then simplify. I get that
    \displaystyle \frac{tan(x) - cot(x)}{tan(x) + cot(x)} = sin^2(x) - cos^2(x)

    Now multiply the top and bottom of this expression by \displaystyle sec^2(x) = \frac{1}{cos^2(x)}. Simplify this a bit and you get your LHS.

    -Dan
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by TacticalPro View Post
    Hi I have a quiz tomorrow and have been looking over some review problems that will be similar to those on the quiz. I've stumbled across a hard one and really need help. I've attempted numerous approaches and still haven't figured it out, so can someone walk me through?

    THANKS! (i removed x's for visual purposes)

    ((tan^2)-1)/(sec^2) = (tan-cot)/(tan+cot)
    Substitute sec^2x=tan^2x+1.
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  4. #4
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    Quote Originally Posted by topsquark View Post
    As usual, I have a way, but I doubt it's the most direct. Take a look at the RHS. Write everything in terms of sines and cosines, add the fractions in the numerator and denominator, then simplify. I get that
    \displaystyle \frac{tan(x) - cot(x)}{tan(x) + cot(x)} = sin^2(x) - cos^2(x)

    Now multiply the top and bottom of this expression by \displaystyle sec^2(x) = \frac{1}{cos^2(x)}. Simplify this a bit and you get your LHS.

    -Dan
    I don't know you get sin^2 - cos^2

    i get (sin^2+cos^2)/(sin^2-cos^2)
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TacticalPro View Post
    I don't know you get sin^2 - cos^2

    i get (sin^2+cos^2)/(sin^2-cos^2)
    \displaystyle \frac{tan(x) - cot(x)}{tan(x) + cot(x)} = \frac{ \frac{sin(x)}{cos(x)} - \frac{cos(x)}{sin(x)} }{ \frac{sin(x)}{cos(x)} + \frac{cos(x)}{sin(x)} }

    Now clear the complex fractions:
    \displaystyle = \frac{ \frac{sin(x)}{cos(x)} - \frac{cos(x)}{sin(x)} }{ \frac{sin(x)}{cos(x)} + \frac{cos(x)}{sin(x)} } \cdot \frac{sin(x)~cos(x)}{sin(x)~cos(x)} = \frac{sin^2(x) - cos^2(x)}{sin^2(x) + cos^2(x)}

    and, of course, the denominator is 1.

    Now multiply by
    \displaystyle \frac{ \frac{1}{cos^2(x)} }{ \frac{1}{cos^2(x)} }

    -Dan
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  6. #6
    Senior Member BAdhi's Avatar
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    here is another way

    use \sec^2x=\tan^2x+1

    so taking LHS

    \displaystyle{\frac{\tan^2x-1}{\tan^2x+1}}

    divide up and down by \tan x
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