⎝ ⎠ ⎝ ⎠ θ =± tan 9 3 cosθ =± 25 5 16 4 169 13 θ θ Note that tan < 0 , so tan 3 θ Note that cosθ must be negative because lies 4 = − in quadrant II. Thus, cosθ 5 . cosθ = 1 = 1 = − 4 13 secθ θ ⎞ sin 13 12 12 − 5 12 ⎛ 5 θ = = tan 4 13 = ⎜ − ⎟ = − θ tan = sin , so cosθ − 13 ⎝ 5 ⎠ 5 θ 13 cosθ = = = cscθ 1 1 13 ( ) (cosθ ) θ θ sin = tan = − 3 ⎛ − 4 ⎞ θ = 3 . sin 12 4 ⎜ 5 ⎟ 5 ⎝ ⎠ secθ = 1 = 1 = − 13 = = = cscθ 1 1 cosθ 5 5 13 θ sin 3 θ cot = 1 = 1 = − 5 θ cot = 1 = 1 = − θ tan 12 4 12 θ tan 3 3
⎝ ⎠ ⎝ ⎠ tan θ =± 9 3 cosθ =± 25 5 16 4 Note that tan θ < 0 , so tan θ 3 4 cosθ = 1 = 1 = − 4 169 13 Note that cosθ must be negative because θ lies in quadrant II. Thus, cosθ = − 5 . 13 secθ − 5 5 sin θ 12 12 ⎛ 13 ⎞ 12 4 tan θ = = 13 = ⎜ − ⎟ = − tan θ = sin θ , so cosθ − 13 13 ⎝ 5 ⎠ 5 cosθ cscθ = 1 = 1 = 13 sin θ = ( tan θ ) (cosθ ) = − 3 ⎛ − 4 ⎞ = 3 . sin θ 12 4 ⎜ 5 ⎟ 5 ⎝ ⎠ secθ = 1 = 1 = − 13 cscθ = 1 = 1 = 5 cosθ 5 13 sin θ 3 cot θ = 1 = 1 = − 5 cot θ = 1 = 1 = − 4 tan θ 12 12 tan θ 3 3
62514-0132318806_ISM_02-1.docx