⎜ ⎟ 5 10 10 ⎝ ⎠ = = θ = = = cot 1 1 100 10 1 1 4 θ tan ⎛ 3 ⎞ 3 cscθ = = = 10 ⎜ ⎟ θ sin ⎝ ⎠ ⎛ 10 ⎞ ⎜ ⎟ ⎝ 10 ⎠ θ = 4 , cosθ < 0, so is in quadrant III 56. θ cot = 1 = 1 = −3 θ cot θ 3 tan ⎛ 1 ⎞ ⎜ ⎟ θ = = = tan 1 1 ⎝ 3 ⎠ 3 ⎜ ⎟ θ cot = − θ > θ 58. secθ 2, tan 0, so is in quadrant III ⎝ ⎠ ⎛ 4 ⎞ 4 θ = + θ Solve for secθ : sec2 1 tan 2 θ θ = + θ Solve for tan : sec2 1 tan 2 =± + secθ 1 tan 2 θ =± θ − tan sec2 1 θ θ Since (− 2)2 − 1 4 − 1 = θ θ tan = sec2 is in quadrant = 3 − 1 = III, secθ < 0 . secθ =− 1 + tan 2 cosθ = 1 1 θ ⎛ 3 ⎞2 9 25 5 secθ 2 =− 1 + ⎜ ⎟ ⎝ 4 ⎠ = − 1 + = − =− θ cot = 1 = 1 ⋅ 3 = 3 16 16 4 cosθ = 1 = − tan 3 3 3 θ 4 secθ 5
⎜ 5 ⎟ 10 10 ⎝ ⎠ cot θ = 1 = 1 = 4 = = 100 10 1 1 tan θ ⎛ 3 ⎞ 3 cscθ = = = 10 ⎜ ⎟ sin θ ⎝ ⎠ ⎛ 10 ⎞ ⎜ ⎟ ⎝ 10 ⎠ 56. cot θ = 4 , cosθ < 0, so θ is in quadrant III cot θ = 1 = 1 = −3 3 tan θ ⎛ 1 ⎞ ⎜ ⎟ tan θ = 1 = 1 = 3 ⎝ 3 ⎠ cot θ ⎛ 4 ⎞ 4 ⎜ ⎟ ⎝ ⎠ 58. secθ = − 2, tan θ > 0, so θ is in quadrant III Solve for secθ : sec2 θ = 1 + tan 2 θ secθ =± 1 + tan 2 θ Solve for tan θ : sec2 θ = 1 + tan 2 θ tan θ =± sec2 θ − 1 Since θ is in quadrant III, secθ < 0 . tan θ = sec2 θ − 1 = (− 2)2 − 1 = 4 − 1 = 3 secθ =− 1 + tan 2 θ cosθ = 1 1 ⎛ 3 ⎞2 9 25 5 secθ 2 =− 1 + ⎜ ⎟ ⎝ 4 ⎠ = − 1 + = − =− 16 16 4 cot θ = 1 = 1 ⋅ 3 = 3 cosθ = 1 = − 4 tan θ 3 3 3 secθ 5
62514-0132318806_ISM_02-1.docx