From the function:
$$H(\omega) = \frac{1}{(1 + j\omega)(1 + j\omega/10)}$$
How is the phase angle obtained when it has multiple poles to get:
$$\phi = -\tan^{-1}(\omega) - \tan^{-1}(\omega/10)$$
What rule of phase angles allows you to separate the two poles into two separate inverse tangent functions?
It is just a matter of manipulating complex numbers.
$$ \angle H(\omega) = \tan^{-1} \left( \frac{\Im\{H(\omega)\}}{\Re\{H(\omega)\}} \right)$$
Where \$\Re \{ \cdot \} \$ is the real part and \$ \Im \{ \cdot \} \$ is the imaginary part. (NOTE: this equality is not always strictly true depending on the signs of the real and imaginary parts of \$H(\omega)\$. When finding the angle of an imaginary number the result may need to be adjusted depending on what quadrant the imaginary number is in.)
Expanding \$ H(\omega) \$ gives
$$ H(\omega) = \frac{1}{-\frac{\omega^2}{10}+\frac{11j\omega}{10}+1} $$
Instead of finding the real and imaginary parts of the whole expression, though you could do that, You can note that:
$$ \angle H(\omega) = \angle \text{numerator of } H(\omega) - \angle \text{denominator of } H(\omega) \\ \angle H(\omega) = \tan^{-1}\left(\frac{0}{1}\right) - \tan^{-1}\left(\frac{\frac{\omega}{10}+\omega}{1-\frac{\omega^2}{10}}\right)\\ $$
Using the arctangent addition,wikipedia, formula the expression can be simplified to
$$ \angle H(\omega) = \phi = -\tan^{-1}(\omega)-\tan^{-1}\left(\frac{\omega}{10}\right) $$
Basically you get a phase contribution term which is the arctangent of each pole location.
