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Math Help - Proving Trig. Identity

  1. #1
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    Proving Trig. Identity

    The original problem is (1-cos(x))/(1+cos(x))=(csc(x)-cot(x))^2

    I start by solving from the right side. First I expand the binomial and then I take reciprocal of both csc and cot. At that point I get ((cos(x) -1)^2)/(1-cos(x)) After that, I have no idea how to get the left side honestly. What do I do from here? Thanks
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  2. #2
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    (\csc{x} - \cot{x})^2 = \left(\frac{1}{\sin{x}} - \frac{\cos{x}}{\sin{x}}\right)^2

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

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

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

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

     = \frac{1 - \cos{x}}{1 + \cos{x}}.
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  3. #3
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    Quote Originally Posted by Altermeris View Post
    The original problem is (1-cos(x))/(1+cos(x))=(csc(x)-cot(x))^2

    I start by solving from the right side. First I expand the binomial and then I take reciprocal of both csc and cot. At that point I get ((cos(x) -1)^2)/(1-cos(x)) After that, I have no idea how to get the left side honestly. What do I do from here? Thanks
    from LHS
    show that: \frac{1-cos(x)}{1+cos(x)}=(csc(x)-cot(x))^2

    \frac{1-cos(x)}{1+cos(x)}=\frac{1-cos(x)}{1+cos(x)} X \frac{1-cos(x)}{1-cos(x)}

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

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

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

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

    \frac{1-cos(x)}{1+cos(x)}=(csc(x)-cot(x))^2

    -- hope it'll help --
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