Question

let 0 be an angle in quadrant IV such that cot0=-(8)/(9). Find the exact value of sin0 and sec0
cot Ø = adjacent/opposite = 8/-9
so let x = 8 and y = -9 in a right-angled triangle in quad IV
r^2 = 64 + 81 = 145
r = √145

sinØ = y/r = -9/√145
secØ = 1/cosØ = r/x = √145/8

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