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Thread: Prove The Equation

  1. #1
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    Prove The Equation

    Given A+B+C=90

    Prove that cotA+cotB+cotC=(cotA)(cotB)(cotC)

    Thanks!
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  2. #2
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    Hello cyt91
    Quote Originally Posted by cyt91 View Post
    Given A+B+C=90

    Prove that cotA+cotB+cotC=(cotA)(cotB)(cotC)

    Thanks!
    Note first that $\displaystyle \cot\theta = \tan(90^o-\theta)$

    So:
    $\displaystyle A+B+C = 90^o$

    $\displaystyle \Rightarrow A = 90^o-(B+C)$

    $\displaystyle \Rightarrow \cot A=\tan(B+C)$, using the above result
    $\displaystyle =\frac{\tan B + \tan C}{1-\tan B \tan C}$

    $\displaystyle =\frac{\cot C+\cot B}{\cot B \cot C - 1}$, multiplying top-and-bottom by $\displaystyle \cot B \cot C$

    $\displaystyle \Rightarrow \cot A \cot B \cot C - \cot A = \cot B + \cot C$

    $\displaystyle \Rightarrow \cot A + \cot B + \cot C = \cot A \cot B \cot C$
    Grandad
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  3. #3
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    Hello, cyt91!

    Another approach . . .


    Given: .$\displaystyle A+B+C\:=\:90^o$

    Prove that: .$\displaystyle \cot\!A+\cot\!B+\cot\!C\:=\:\cot\!A\cot\!B\cot\!C$
    We need this identity: .$\displaystyle \cot(\alpha + \beta) \:=\:\frac{\cot\alpha\cot\beta - 1}{\cot\alpha + \cot\beta}$

    Also note that: .$\displaystyle \cot(90^o - \alpha) \;=\;\tan\alpha \;=\;\frac{1}{\cot\alpha} $



    We have: - - . $\displaystyle A + B \;=\;90^o - C$

    Then: . . . $\displaystyle \cot(A + B) \;=\;\cot(90^o - C)$

    . . . . . .$\displaystyle \frac{\cot\!A\cot\!B - 1}{\cot\!A + \cot\!B} \;=\;\frac{1}{\cot\!C} $

    . .$\displaystyle \cot\!C(\cot\!A\cot\!B - 1) \;=\;\cot\!A + \cot\!B$

    $\displaystyle \cot\!A\cot\!B\cot\!C - \cot\!C \;=\;\cot\!A+ \cot\!B$

    . .$\displaystyle \cot\!A + \cot\!B + \cot\!C \;=\;\cot\!A\cot\!B\cot\!C$



    Nice explanation, Grandad!
    .
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  4. #4
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    Thanks a lot. You guys have been helpful. Keep up the good work!
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