Inverse Trigonometric functions question I have problem - I have a solution also, though. But still a confusion.
I have to evaluate the expression :  $arctan \sqrt{\dfrac{x(x+y+z)}{yz}} + arctan \sqrt{\dfrac{y(x+y+z)}{xz}} + arctan \sqrt{\dfrac{z(x+y+z)}{xy}}$
Now, I observed that $\sqrt{\dfrac{x(x+y+z)}{yz}}(=A)+ \sqrt{\dfrac{y(x+y+z)}{xz}}(=B) + \sqrt{\dfrac{z(x+y+z)}{xy}}(=C) = ABC.$ Hence, $A+B+C=ABC.$ Now,When this happens, $\arctan x + \arctan y + \arctan z= 0$ or $\pi$.

Now, My question is, what is the answer of my question, $0$ or $\pi$ ? If both answers are correct then for which conditions under the answer willl be correct. 

We know the principal value of $\arctan$ lies in $\in\left[-\dfrac\pi2,\dfrac\pi2\right]$
As $\sqrt a\ge0$ for real $a$
$\sqrt{\dfrac{x(x+y+z)}{yz}}\ge0$ etc.
and we need all the terms under root to be non-negative for the real argument of $\arctan$
$\implies$ the principal value of $\arctan\sqrt{\dfrac{x(x+y+z)}{yz}}$ will lie in $\in\left[0,\dfrac\pi2\right]$
$\implies$ the value of $\sum\arctan\sqrt{\dfrac{x(x+y+z)}{yz}}$ will lie in $\in\left[0,\dfrac{3\pi}2\right]$
Case$\#1: x=y=z=c$
Case$\#1A:$ What if $c\ne0$
Case$\#1B:$ What if $c=0$
Case$\#2:$ Two of them equal $0$
Case$\#3:$ One of them equals $0$
Case$\#4:$ If $xyz\ne0,\sum\arctan\sqrt{\dfrac{x(x+y+z)}{yz}}>0$
$\sum\arctan\sqrt{\dfrac{x(x+y+z)}{yz}}=\pi$
