unit circle problem - tan ratio in the third quadrant

Question

I am convinced that the answer is E because as $x<0$ and $y>0$, therefore $\tan(\theta)=\Bigg|\dfrac{-y}{x}\Bigg|=\dfrac{y}{-x}$.

Can anyone please confirm?

Thank you!

Answer 1

If the point that corresponds to the angle $\theta$ has coordinates $P(\theta) = (x,y)$, then we define $\tan \theta = \dfrac yx$.

So if $P(\theta) = (x,-y)$, then $\tan \theta = \dfrac {-y}{x} = -\dfrac yx$.