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Math Help - tangent and cotangent

  1. #1
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    tangent and cotangent

    can someone explain this to me:

    tan(theta) = 60/x

    x = 60/tan(theta)

    x = 60cot(theta)

    i dont understand how the tangent switches to cotangent.
    Last edited by algebra2; February 18th 2008 at 01:27 PM. Reason: Invalid error.
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by algebra2 View Post
    can someone explain this to me:

    tan(theta) = 60/x

    x = 60/tan(theta)

    x = 60cot(theta)

    i dont understand how the tangent switches to cotangent.
    Cotangent is just 1 over tangent so
    cot(x) = \frac 1{tan(x)}

    In the same way that cosecant is 1 over sine, and secant is 1 over cosine.

    --------------------------------------------------

    So your intitial equation is:
    tan(\theta)=\frac{60}x

    then multiply both sides by \frac{x}{tan(\theta)}
    \frac{x}{tan(\theta)}*tan(\theta)=\frac{60}x*\frac  {x}{tan(\theta)}

    Simplify (cancel out tangents on LHS and x's on RHS) to get your second equation.
    x=\frac{60}{tan(\theta)}

    You can partition the RHS like so:
    x=60*\frac{1}{tan(\theta)}

    And you see that on the RHS 1/tangent = cotangent so
    x=60*cot(\theta)

    Which is your last equation.

    --------------------------------------------------

    Some other things to note:

    since cot(\theta) = \frac 1{tan(\theta)} you can multiply both sides by tangent over cotangent and get tan(\theta)=\frac 1{cot(\theta)}

    and also since tan(\theta) = \frac{sin(\theta)}{cos(\theta)}

    it follows that cot(\theta) = \frac 1{tan(\theta)} = \frac{1}{\frac{sin(\theta)}{cos(\theta)}} = \frac{cos(\theta)}{sin(\theta)}

    so cot(\theta)=\frac{cos(\theta)}{sin(\theta)}
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