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sin(t) = 2/3

sin2(t) + cos2(t) = 1

so cos(t) = ± sqrt[1 - sin2(t)]

but because t is in the first quadrant, cos(t) > 0 so cos(t) = + sqrt[1 - sin2(t)]

= sqrt[1 - 4/9] = sqrt[5/9] = sqrt(5)/3

Then sec(t) = 1/cos(t) = 1/sqrt(5)/3 = 3/sqrt(5) = 3*sqrt(5)/5

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Q: What is sec theta if sin theta equals 2 over 3 with theta in quadrant 1?
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