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Thread: Another tan sum problem!

  1. #1
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    help..

    tan(A-B)+tan(B-C)+tan(C-A)=tan(A-B)tan(B-C)tan(C-A)

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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Joyce
    help..

    tan(A-B)+tan(B-C)+tan(C-A)=tan(A-B)tan(B-C)tan(C-A)

    tan(A-B)= -tan(B-A) =-tan((B-C)+(C-A))
    expand it and you will get the answer
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Joyce
    help..

    tan(A-B)+tan(B-C)+tan(C-A)=tan(A-B)tan(B-C)tan(C-A)

    This is a new question it belongs in a new thread.

    Also don't post the same question twice.

    RonL
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  4. #4
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    Hello, Joyce!

    $\displaystyle \tan(A-B) + \tan(B-C) + \tan(C-A)\;=\;\tan(A-B)$$\displaystyle \tan(B-C)\tan(C-A)$
    Let $\displaystyle P = A-B,\;\;Q = B-C,\;\;R = C-A$

    We see that: $\displaystyle P + Q + R \:= \A - B) + (B - C) + (C - A)\:=\:0$

    So we have: $\displaystyle P + Q + R\:=\:0\quad\Rightarrow\quad P + Q\:=\:-R$


    Take the tangent of both sides: $\displaystyle \tan(P + Q)\;=\;\tan(-R)$

    . . and we have: $\displaystyle \frac{\tan(P) + \tan(Q)}{1 - \tan(P)\cdot\tan(Q)}\;=\;-\tan(R)$

    Then: $\displaystyle \tan(P) + \tan(Q)\;=\;-\tan(R)[1 - \tan(P)\tan(Q)] \;=\;$$\displaystyle -\tan(R) + \tan(P)\tan(Q)\tan(R) $

    . . Hence: $\displaystyle \tan(P) + \tan(Q) + \tan(R)\;=\;\ran(P)\tan(Q)\tan(R)$


    Therefore: $\displaystyle \tan(A-B) + \tan(B-A) + \tan(C-A) \;=\;\tan(A-B)$$\displaystyle \tan(B - C)\tan(C - A)$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    How did I come up with this proof?

    I saw that this problem was similar to another (classic) problem:
    . . If $\displaystyle A + B + C = 180^o$, then: $\displaystyle \tan A + \tan B + \tan C\;=\,\tan A\tan B\tan C$

    And the proof is almost identical.
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