Question 398188
Suppose sec alpha = -7/4 and that the terminal side of alpha is in quadrant III. find sin alpha and tan alpha. 
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sec(theta) = r/x
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If theta is in the 3rd Quadrant, x is negative.
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So, x = -4, r = 7
Solve for "y", which is negative in QIII
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y^2 = r^2-x^2
y^2 = 49-16
y^2 = 33
y = -sqrt(33)
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sin(alpha) = y/r = -sqrt(33)/7
tan(alpha) = y/x = -sqrt(33)/-4 = sqrt(33)/4
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cheers,
Stan H.