Question 901453
sec of theta= -6 given that sin theta is > 0. find the remaining six trigonometric functions of theta
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Since sec is negative and sin is positive theta is in QII
By definition sec = r/x
Since sec(t) = -6, r = 6 and x = -1
Then y = sqrt[r^2-x^2[ = sqrt(36-1] = sqrt(35)
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Ans:: sin(t) = y/r = sqrt(35)/6
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Cheers,
Stan H.