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Math Help - 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 \cot\theta = \tan(90^o-\theta)

    So:
    A+B+C = 90^o

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

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

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

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

    \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: .  A+B+C\:=\:90^o

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

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



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

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

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

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

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

    . . \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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