If $a, b, c$ are real positive numbers and
$\theta=\tan ^{-1}\left[\frac{a(a+b+c)}{b c}\right]^{\frac{1}{2}}+\tan ^{-1}\left[\frac{b(a+b+c)}{c a}\right]^{\frac{1}{2}}+ \tan ^{-1}\left[\frac{c(a+b+c)}{a b}\right]^{\frac{1}{2}}$
then $\tan \theta$ equals?
I did this by adding the arctan terms twice, It took a lot of time and a ton of trials and error. I don't even want to type the solution in TeX
I know this is a symmetric equation, which are usually easy to solve (Still learning that) but I'm not able to simplify it. How can I solve it in a more simple way?
(And help me with the tags too)
It would be nice to first the simplify the argument of $\tan^{-1}$
Let $ \dfrac{a+b+c}{abc}=s^2$ so the sum looks simple $$\theta= \tan ^{-1}(as )+\tan ^{-1}(bs) +\tan ^{-1}(cs )$$
Now you can do the same, or simply use the identity:
If $x_1,x_2 \ldots x_n$ are real then $$\tan ^{-1} x_{1}+\tan ^{-1} x_{2}+\ldots+\tan ^{-1} x_{n}=\tan ^{-1}\left(\frac{S_{1}-S_{3}+S_{5}-S_{7}+\ldots}{1-S_{2}+S_{4}-S_{6}+\ldots}\right)$$ Where $S_{k}$ denotes the sum of products of $x_{1}, x_{2}, \ldots x_{n}$ taken $k$ at a time.
And in this equation, $ \displaystyle \sum_{cyc} \tan^{-1}(s\cdot a_i) $
$$\theta = \tan ^{-1}\left(\frac{(as+bs+cs) - (abcs^3)}{1-(a bs^{2}+bcs^{2}+a cas^{2})} \right) \\ = \tan ^{-1}\left(\frac{s\left((a+b+c)-a b c s^{2}\right)}{1-s^2(ab+b c+a c)}\right)$$
See that $ \left( a+b+c - abc \cdot s^2 \right) = 0$
Thus, $\tan (\theta) = 0$