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Math Help - simplifying a complex fraction

  1. #1
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    simplifying a complex fraction

    h=(Tanά)d x tanβx
    Tanά-tanβ
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cocoknny View Post
    h=(Tanά)d x tanβx
    Tanά-tanβ
    Is this
    h = \frac{tan(\alpha)}{tan(\alpha) - tan(\beta)} \times tan(\beta x)

    If not your expression is too unclear to work with. Could you possibly write it out and scan it?

    -Dan
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  3. #3
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    hello

    hello, um what u posted is right um i didn't have the functions u have on ur computer to do mine like yours. If u can assist in solving that fraction in anyway it is greatly appreciated.
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  4. #4
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    hello

    in this one i have to change tan to cotangent.
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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by cocoknny View Post
    in this one i have to change tan to cotangent.
    I can't tell what your equation is, but I hope this formula will help

    This is the subtraction formula for tangent
    tan(x-y) = \frac{tan(x) - tan(y)}{1+tan(x)tan(y)}

    If yours is the cotangent, it should be
    \frac{1}{tan(x-y)} = \frac{1}{\frac{tan(x) - tan(y)}{1+tan(x)tan(y)}}

    or
    cot(x-y) = \frac{1+tan(x)tan(y)}{tan(x) - tan(y)}

    because cot(x) = \frac{1}{tan(x)}
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cocoknny View Post
    hello, um what u posted is right ...
    if so, what was thar "d" after \tan \alpha that you had in your problem that topsquark did not write in his? please double check to make sure you have the correct thing, we don't want to tackle a problem only to find out that we solved the wrong one
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  7. #7
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    Hello, cocoknny!

    If that's really \beta x, we've got trouble ... with a capital T.


    Simplify:. h\;=\;\frac{\tan\alpha\tan\beta}{\tan\alpha - \tan\beta}
    It doesn't really simplify . . . we can only exchange it for another expression.


    We have: . \Large{\frac{\frac{\sin\alpha}{\cos\alpha}\cdot\fr  ac{\sin\beta}{\cos\beta}}{\frac{\sin\alpha}{\cos\a  lpha} - \frac{\sin\beta}{\cos\beta}}}

    Multiply top and bottom by \cos\alpha\cos\beta\!:\;\;\frac{\sin\alpha\sin\bet  a}{\sin\alpha\cos\beta - \sin\beta\cos\alpha}

    . . And we have: . \frac{\sin\alpha\sin\beta}{\sin(\alpha - \beta)}

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