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Math Help - Help required on solving Trigo sum..

  1. #1
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    Help required on solving Trigo sum..

    Prove the identity

    [1+(tanA)^2] / [1-(tanA)^2] = sec2A

    Thanks in advance
    Last edited by Punch; December 26th 2009 at 06:15 AM.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Punch View Post
    Prove the identity

    [1+(tanA)^2] / [1-(tanA)^2] = sec2A

    Thanks in advance
    Start with the LHS

    1+tan^2A = sec^2A

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

    cos^2A cancels leaving \frac{1}{cos^2A-sin^2A} = \frac{1}{cos(2A)} = sec(2A) = RHS
    Last edited by e^(i*pi); December 26th 2009 at 05:55 AM. Reason: tidying up latex
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Start with the LHS

    1+tan^2A = sec^2A

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

    cos^2A cancels leaving \frac{1}{cos^2A-sin^2A} = \frac{1}{cos(2A)} = sec(2A) = RHS
    sorry but i only understood the part 1+tan^2A = sec^2A

    this may provide a better picture, [1+(tanA)^2] / [1-(tanA)^2] = sec2A

    it's a fraction
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  4. #4
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    so since 1+tan^2A = sec^2A

    For [1 + (tanA)^2] / [1 - (tanA)^2] = sec2A, starting with LHS,

    [1 + (tanA)^2] / [1 - (tanA)^2] = sec^2A / [1 - (tanA)^2]
    how do i continue?
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  5. #5
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    \frac{1+\tan^2{A}}{1-\tan^2{A}} \cdot \frac{\cos^2{A}}{\cos^2{A}} = \frac{\cos^2{A} + \sin^2{A}}{\cos^2{A}-\sin^2{A}} = \frac{1}{\cos(2A)} = \sec(2A)
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  6. #6
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Punch View Post
    sorry but i only understood the part 1+tan^2A = sec^2A

    this may provide a better picture, [1+(tanA)^2] / [1-(tanA)^2] = sec2A

    it's a fraction
    Yeah, I was working on the left side to get the right. Just found it easier to simplify the numerator first.

    Quote Originally Posted by Punch View Post
    so since 1+tan^2A = sec^2A
    Yeah (you can derive this from sin^2A+cos^2A=1)

    For [1 + (tanA)^2] / [1 - (tanA)^2] = sec2A, starting with LHS,

    [1 + (tanA)^2] / [1 - (tanA)^2] = sec^2A / [1 - (tanA)^2]
    how do i continue?
    Convert your tan^2 into sin and cos

    1-tan^2A = 1 - \frac{sin^2A}{cos^2A}

    Then you'll want to make this into one fraction and since the LCM is cos^2A we'll use that. \left(1 = \frac{cos^2A}{cos^2A}\right)

    \frac{cos^2A}{cos^2A} - \frac{sin^2A}{cos^2A}

    As these have the same denominator we can combine fractions so the above is equal to \frac{cos^2A-sin^2A}{cos^2A}

    You should know that from your double angle identities that cos^2A-sin^2A = cos(2A) and you should also know that when dividing by a fraction we flip it and multiply.

    \frac{sec^2A}{\frac{cos(2A)}{cos^2A}} = \frac{sec^2A}{cos(2A)} \times cos^2A

    From the definition of sec: sec^2A \times cos^2A = 1 so that cancels to leave \frac{1}{cos(2A)} and again from the definition of sec:

    \frac{1}{cos(2A)} = sec(2A)

    Since this is equal to the RHS we've proved the identity is true
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