Question 1148549
Equation of the line: 3x + 5y = 0 i.e. y = -(3/5)x

Slope of the line = -3/5

tan θ = -3/5 (negative since in Quadrant IV)
cot θ = 1/tan θ = 5/3


We know
cosec² □ - cot² θ = 1
cosec² θ = 1 + cot² θ
cosec θ = √(1 + cot² θ) = -√(1 + (5/3)²) = -√34/3

sin θ = 1/cosec θ = -3/√34

cos θ = √(1 - sin² θ) = √(1 - 9/34) = 5/√34
sec θ = 1/cos θ = √34/5