SOLUTION: Let θ be an angle in quadrant III, such that csc θ= {{{-9/8}}} Find tan θ & cos θ: So far, I found that cosθ = {{{-8/9}}}. As for tanθ, I think

Algebra ->  Trigonometry-basics -> SOLUTION: Let θ be an angle in quadrant III, such that csc θ= {{{-9/8}}} Find tan θ & cos θ: So far, I found that cosθ = {{{-8/9}}}. As for tanθ, I think      Log On


   

Question 985362: Let θ be an angle in quadrant III, such that csc θ= -9%2F8
Find tan θ & cos θ:
So far, I found that cosθ = -8%2F9.
As for tanθ, I think it is +8sqrt%28+17+%29%2F%289%29+
I got that through -9%5E2+=+8%5E2+%2B+x%5E2 with x= sqrt%2817%29
And using that, I saw that tanθ's equation would be %28sqrt%2817%29%2F%28-9%29%29 all over -8%2F9
Right up until I finished typing up this question, I just realized that I misused the reciprocals, oops. I would still like to know the proper answer.
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
csc theta = -9/8
note that csc theta = 1 / sin (theta), therefore
1 / sin theta = -9 / 8
sin theta = -8 / 9
inverse sin theta = −62.733955549 approx -62.7
theta is in quadrant III, theta = 270 - 62.7 = 207.3 degrees
*********************************************************************
cos 207.3 = −0.888617233
tan 207.3 = 0.516138488