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Thread: Prove(2)

  1. #1
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    Prove(2)

    Prove $\displaystyle
    \frac{1+sec2A}{tan2A}=cotA
    $
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  2. #2
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    Quote Originally Posted by Punch View Post
    Prove $\displaystyle
    \frac{1+sec2A}{tan2A}=cotA
    $
    $\displaystyle \frac{1+sec2\alpha}{tan2\alpha} = \frac{1 + \frac{1}{cos(2\alpha)}}{\frac{sin(2\alpha)}{cos(2\ alpha)}} = \frac{cos(2\alpha)+1}{cos(2\alpha)} \cdot \frac{cos(2\alpha)}{sin(2\alpha)}
    =$

    $\displaystyle = \frac{cos(2\alpha)+1}{sin(2\alpha)} = \frac{cos^2(\alpha) - sin^2(\alpha) + cos^2(\alpha)+sin^2(\alpha)}{2sin(\alpha)cos(\alph a)} = \frac{2cos^2(\alpha)}{2sin(\alpha)cos(\alpha)} = $ $\displaystyle \frac{cos(\alpha)}{sin(\alpha)} = cot(\alpha)$

    Identities used were:

    $\displaystyle (1)~ cos(2\alpha) = cos^2(\alpha)-sin^2(\alpha)$
    $\displaystyle (2) ~sin(2\alpha) = 2sin(\alpha)cos(\alpha)$
    $\displaystyle (3) ~1 = sin^2(\alpha)+cos^2(\alpha)$

    These really are not difficult... simply expand and use a list of trigonometric identities
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    $\displaystyle \frac{1+sec2\alpha}{tan2\alpha} = \frac{1 + \frac{1}{cos(2\alpha)}}{\frac{sin(2\alpha)}{cos(2\ alpha)}} = \frac{cos(2\alpha)+1}{cos(2\alpha)} \cdot \frac{cos(2\alpha)}{sin(2\alpha)}
    =$

    $\displaystyle = \frac{cos(2\alpha)+1}{sin(2\alpha)} = \frac{cos^2(\alpha) - sin^2(\alpha) + cos^2(\alpha)+sin^2(\alpha)}{2sin(\alpha)cos(\alph a)} = \frac{2cos^2(\alpha)}{2sin(\alpha)cos(\alpha)} = $ $\displaystyle \frac{cos(\alpha)}{sin(\alpha)} = cot(\alpha)$

    Identities used were:

    $\displaystyle (1)~ cos(2\alpha) = cos^2(\alpha)-sin^2(\alpha)$
    $\displaystyle (2) ~sin(2\alpha) = 2sin(\alpha)cos(\alpha)$
    $\displaystyle (3) ~1 = sin^2(\alpha)+cos^2(\alpha)$

    These really are not difficult... simply expand and use a list of trigonometric identities
    Sometimes, I am really stucked at proving, I have a phobia of it and was stucked when I got to \frac{cos2A-1}{sin2A}... Seems like I need more practise in proving, give me tips/comments if you have any
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