How do I get the arctan-addition-rule to work with a multiplication infront of one of the terms? i need help with this. I think, I'm missing a simple rule here, but i don't get it:
$180^\circ = \arctan(10 \omega) + 2 \arctan(\omega)$
I know that $\omega$ will be $1.099$ but I don't get how.
 A: Hint: Think of it as "$\arctan 10\omega + (\arctan\omega +\arctan\omega)$" and deal with the part in parentheses first. Then do it again since it will then be of the form "$\arctan 10\omega + \arctan(\textrm{first result})$".
A: We have $\arctan(10\omega)=2\left(90^\circ-\arctan\omega\right)=2\text{arccot}\omega$
Now using Principal values of Inverse trigonometric functions,
$-\dfrac\pi2<\arctan(10\omega)<\dfrac\pi2\implies-\dfrac\pi4<\text{arccot}\omega<\dfrac\pi4$
But $0<\text{arccot}\omega<\pi\implies0<\text{arccot}\omega<\dfrac\pi4\iff\dfrac\pi4<\arctan x<\dfrac\pi2$
$\implies1<\omega<\infty\  \ \ \  (1)$
$\iff0<\dfrac1\omega<1\  \ \ \  (2)$
Using Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?, $$\text{arccot}\omega=\arctan\dfrac1\omega$$
Using my answer here Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$  by $(2)$
$$2\arctan\dfrac1\omega=\cdots=\arctan\dfrac{2\omega}{\omega^2-1}$$
So, we have $$10\omega=\dfrac{2\omega}{\omega^2-1}$$
Now we must honor $(1)$
