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$\tan(\theta) = -\frac{15}{8}$ 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.

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    Sorry the question was in the title, I need to find the trig function values.2011-06-02

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You can compute $r=\sqrt{x^2+y^2}=\sqrt{(-8)^2+15^2}=17$ and evaluate the trig functions as in this answer by user6312 to a question of yours.

Added: In response to your comment. Edited: For the remaing trigonometric functions you have, by definition, the following five fractions:

  • $\cos\theta=\frac{x}{r}$,
  • $\sin\theta=\frac{y}{r}$,
  • $\cot\theta=\frac{1}{\tan\theta}=\frac{x}{y}$,
  • $\sec\theta=\frac{1}{\cos\theta}=\frac{r}{x}$,
  • $\csc\theta=\frac{1}{\sin\theta}=\frac{r}{y}$.

To find their values you just have to substitute $x=-8$, $y=15$ (found by you) and $r=17$ (evaluated above) in these expressions.

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    @Adam: I added and edited my answer.2011-06-02
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Draw a picture! You can then get all six triggies easily.

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    @Adam, did you see the question, "At what angle do those two sides meet?" Do you know what people call that kind of triangle? Do you know anything about the lengths of the sides of that kind of triangle?2011-06-02