Asked by JOHNNY

let theta be an angle of quadrant IV such that sin theta = -8/9 find the exact value of sec theta and cot theta?
Answered by R_scott
sin(Θ) = -8/9 = y/r

x = √(r^2 - y^2) = √17

sec(Θ) = r / x

cot(Θ) = x / y
Answered by Damon
third leg along x axis is sqrt (81-64) =sqrt (17)
so
x = sqrt (17)
y = -8
hypotenuse = 9
cos = sqrt (17) / 9
so sec = 1/cos = 9/sqrt 17
cot = sqrt (17) / -8
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