Inverse tangents in cyclic order

In summary, the problem asks to find the value of tan(theta) if theta is equal to the inverse tangent of (a(a+b+c)/bc) plus the inverse tangent of (b(a+b+c)/ac) plus the inverse tangent of (c(a+b+c)/ab). Attempts at solving the problem involve considering the sides of a triangle and using their properties, but ultimately, the answer is found by using the equation for the sum of inverse tangents and solving for tan(theta) using the tangent sum formula.
  • #1

Homework Statement

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 $$tan\theta$$

Homework Equations

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!
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  • #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.
  • #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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