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I've got this function in Fourier space: $G(j\omega)=K\cdot\frac{1-j\omega T}{1+j\omega T}$ $\qquad K,T$ are real and positive

I'd like to have a phase plot, so I try to get the argument of my complex number/function: $G(j\omega)=K\cdot\frac{(1-j\omega T)\cdot (1-j\omega T)}{(1+j\omega T)\cdot (1-j\omega T)}=K\cdot\frac{-T^2\omega^2+1-j2T\omega}{1+\omega^2T^2}$

$\DeclareMathOperator{\arg}{arg} \arg{(G(j\omega))}:=\phi=\arctan{\left(\frac{-2T\omega}{1-T^2\omega^2}\right)} $

The sample solution says with no comment: $\phi=-2\arctan{(\omega T)}$

What am I doing wrong? Or is there any way to convert between both solutions?

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HINT:

Use the tangent double angle formula to find $\tan(-2\arctan(\omega T))$. Compare the result to the tangent of your answer.

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