1.) sec t = 3, terminal point of t is in quadrant 4
2.) sin t = 1/4, sec t < 0
Please show the steps… thanks!
2 Answers

1)
sec t = 3 ==>hypotenuse / adj. side = 3/1
hypotenuse = 3, adj. side = 1
opposite side = √(9 – 1) = 2√2
in IV quadrant only cos and sec are positive
sin t= – 2√2 / 3 , csc t = – 3 / (2√2)
cos t = 1/3
tan t = 2√2, cot t = 1 /(2 √2)
2)
sin t = 1/4===> opposite side = 1, hypotenuse = 4
adj. side = √(4^2 – 1) = √15
since sec and sin are negative, t is in III quadrant, where only tan and cot are positive.
sin t = 1/4, csc t = 4
cos t = √15 / 4, sec t = 4 /√15
tan t = 1/√15, cot t = √15

cost=4/5=> t=arc cos(4/5)=> t=143* 8′ or 216*fifty two’eleven” simply by fact cost is ve interior the II & III quadrants. Now attitude t is had to terminate interior the III quadrant, so the respond is t=216* fifty two’ eleven”.