$\tan(\theta) = -\frac{15}{8}$ given that $\theta$ is in quadrant II
I know that $x= -8$ and $y= 15$ since it is in quadrant II $x$ has to be the negative. Where do I go from here? I tried $\tan^2\theta - \sec^2\theta = 1$ got some nonsensical answers.
Not sure how the $-\frac{15}{8}$ functions either, bad math on my part I know but I don't know if it squared is positive or negative. I mean logically to me it is positive but I am not sure, I can't get a proper answer either way.