tan(θ + ϕ);

Question

cos(θ) = − 1/3 θ in Quadrant III, sin(ϕ) = 1/4
ϕ in Quadrant II

evaluate the expression

Answer 1

θ = 180 + cos^-1 (1/3)
= 180 + 70.5
= 250.5

ϕ = 180 - sin^-1 (1/4)
= 165.5

so we want tan (416)
= 1.48

Answer 2

cos(θ) = −1/3 QIII, so tanθ = √8
sin(ϕ) = 1/4 in QII, so tanϕ = -√15/4

tan(θ+ϕ) = (tanθ + tanϕ)/(1-tanθ tanϕ)
= (√8 - √15/4)/(1-(√8)(-√15/4))
= (18√15 - 31√2)/52

Answer 3

oops. tanϕ = -1/√15

tan(θ+ϕ) = (√8 - 1/√15)/(1-(√8)(-1/√15))
= (32√2 - 9√15)/7
= 1.48