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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.

Jo Schi
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    If you can deal with $\arctan x+\arctan y$, then you can also deal with $\arctan x+\arctan y+\arctan y$. –  Sep 29 '17 at 14:42

2 Answers2

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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 "$\arctan 10\omega + \arctan(\textrm{first result})$".

MPW
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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 $(1)$

lab bhattacharjee
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