Nyquist plot Phase Question
\$L(j\omega)=\frac{K(-j-T_1\omega)}{\omega(1+T_1^2\omega^2)}\$

Here the coordinate comes in the third quadrant so you need to take 

\$\theta=\pi+\tan^{-1}(\frac{1}{T_1\omega})\$

so at \$w\to0\$,\$\theta=180+90=270=-90\$

and at \$w\to\infty\$,\$\theta=180+0=180\$