<PRE><B>Question 1032166</B>
 If csc(theta)= -5/3 and theta has its terminal side in Quadrant III, find the exact value of tan(2theta). 
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csc = r/y = -5/3 in QIII where x and y are negative.
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So r = 5 and y = -3
Then x = -sqrt(r^2-y^2) = -sqrt(16) = -4
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tan(t) = y/x = 3/4
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tan(2*t) = 2tan(t)/[1-tan^2(t)] = 2(3/4)/[1-(9/16)] = (3/2)/(7/16) = 24/7
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Cheers,
Stan H.
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