How do I find the phase of a transfer function in different ways?
Number 1: In your original expression there is no minus sign. It suddenly appears when you start calculating phase angles.
Number 2: All those expressions are equivalent and do yield the same phase. Your error is in misinterpreting the minus sign. Your first calculation (in black) is correct. Note that if there had been no minus sign, the first term would be positive which is a difference of π or 180 degrees which is the effect of multiplying by -1. In your second equation (the first red one), the minus sign again creates a phase of -π (not multiplying the phase by -1, which, when added to your expression, will result in the same expression as your first calculation. In your third calculation, moving the -1 to the denominator , adds a phase of π to the expression, and also results in the same result as your first equation. I will leave it as an exercise to figure out the correct interpretation of the minus signs.