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Inverse tangents in cyclic order

  1. May 31, 2016 #1
    1. The problem statement, all variables and given/known data

    The problem- if

    $$\theta= tan^{-1}(\frac{a(a+b+c)}{bc})+tan^{-1}(\frac{b(a+b+c)}{ac})+tan^{-1}(\frac{c(a+b+c)}{ab})$$
    , then find $$tan\theta$$

    2. Relevant equations


    3. The attempt at a solution
    I tried to use these as sides of a triangle and use their properties, but other than that I am clueless. I cannot think of a substitution either. The answer happens to be zero. Any help is appreciated. Thanks in advance!!
     
  2. jcsd
  3. May 31, 2016 #2
    Okay, I have it. Apparently, there is no 'elegant' way to do this. Just use the equation for $$arctan x + arctan y + arctan z$$, and after some really messy calculation, you get the answer as zero. If there is a different, neat answer, please let me know.
     
  4. Jun 2, 2016 #3
    i think your question is wrong it should be:-
    in all the terms it should be sqrt(x) , sqrt(y), sqrt(z) inside of arctan ; where x,y,z are a(a+b+c)/bc , b(a+b+c)/ac , c(a+b+c)/ab ; as i have seen a similar question in my textbook.

    if you consider it like this and take tan both sides and use tan(E+F+G)=(S1-S3)/(1-S2), it is easily solved in just few steps.
     
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