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Math Help - counting by two methods

  1. #1
    Super Member dhiab's Avatar
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    counting by two methods

    reals numbers.
    1 ) Calculate :

    2 ) Let :

    Calculate by anathor mhetode the real number E .
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  2. #2
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    Gello, dhiab!

    I hope I understand the problem . . .

    We should know this identity: . \tan(\alpha - \beta) \;=\;\frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}


    Given: . \begin{array}{c}a \:=\:\tan A \\ b \:=\:\tan B \\ x \:=\:\tan X\end{array}

    Calculate: . E \;=\;\frac{\dfrac{x-a}{1+ax} - \dfrac{x-b}{1+bx}} {1 + \dfrac{x-a}{1+ax}\cdot\dfrac{x-b}{1+bx}}

    We have: . E \;=\;\frac{\dfrac{\tan X - \tan A}{1 + \tan A\tan X} - \dfrac{\tan X - \tan B}{1 + \tan B\tan X} }{1 + \dfrac{\tan X - \tan A}{1 + \tan A\tan X}\cdot\dfrac{\tan X - \tan B}{1 + \tan B\tan X}}


    . . . . . . . . E \;=\;\frac{\tan(X-A) - \tan(X-B)}{1 + \tan(X-A)\tan(X-B)}


    . . . . . . . . E \;=\;\tan\bigg[(X-A) - (X-B)\bigg]


    . . . . . . . . E \;=\;\tan(B-A)

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