θ θ and cot < 0 , so lies in quadrant = − 5 34. θ and cosθ < 0 , so lies in quadrant III. 3 θ cot = 12 36. 5 Using the Pythagorean Identities: IV. θ θ csc2 = 1 + cot 2 Using the Pythagorean Identities: 2 2 θ csc2 = 1 + ⎛ 12 = 1 + 144 = cot θ θ = csc − 1 ⎞ 169 2 ⎜ ⎟ 25 25 − = − 1 25 1 θ = 16 cot = ⎛ − 5 ⎞ 2 ⎜ ⎟ 3 9 9 cscθ =± 169 13 ⎝ ⎠ 25 5 θ θ cot =± 16 = ± 4 Note that cscθ must be negative because lies in 9 3 quadrant III. Thus, cscθ 13 θ θ Note that cot < 0 , so cot 4 5 3 θ sin = 1 = 1 = − 5 θ sin = 1 = 1 = − 3 cscθ 13 13 cscθ 5 5 θ cot = cosθ , θ cot = cosθ , θ so sin θ so sin (cot ) (sin ) θ θ cosθ = = 12 ⎛ − 5 ⎞ = (cot ) (sin ) θ θ cosθ = 4 ⎛ − 3 ⎞ = 4 − 12 5 ⎜ 13 ⎟ 13 3 ⎜ 5 ⎟ 5 ⎠ ⎝ ⎠ θ θ tan = 1 = 1 = 5 tan = 1 = 1 = − 3 ⎝ θ cot θ cot 4 12 4 12 secθ = 1 = 1 = − 13 secθ = 1 = 1 = 5 cosθ 12 12 cosθ 4 37. θ θ = 12 and lies in quadrant II. 35. θ θ and tan < 0 , so lies in quadrant II. 13 secθ 5 4 sin Using the Pythagorean Identities: Using the Pythagorean Identities: θ θ tan = sec2 −1 2 θ θ cos2 = 1 − sin 2 2 2 θ = ⎛ − tan 5 2 θ ⎞ − 1 = 25 − 1 = cos2 = 1 − ⎛ 12 = 1 − 144 = 25 9 ⎞ ⎜ 4 ⎟ 16 16 169 169 ⎜ ⎟ 13
34. cot θ = 12 5 and cosθ < 0 , so θ lies in quadrant III. 36. = − 5 3 and cot θ < 0 , so θ lies in quadrant Using the Pythagorean Identities: csc2 θ = 1 + cot 2 θ IV. Using the Pythagorean Identities: 2 2 csc2 θ = 1 + ⎛ 12 ⎞ = 1 + 144 = 169 cot θ = csc θ − 1 2 ⎜ ⎟ 25 25 cot 2 θ = ⎛ − 5 ⎞ − 1 = 25 − 1 = 16 ⎜ 3 ⎟ 9 9 cscθ =± 169 13 25 5 Note that cscθ must be negative because θ lies in quadrant III. Thus, cscθ 13 5 sin θ = 1 = 1 = − 5 ⎝ ⎠ cot θ =± 16 = ± 4 9 3 Note that cot θ < 0 , so cot θ 4 3 sin θ = 1 = 1 = − 3 cscθ 13 13 cscθ 5 5 cot θ = cosθ , so sin θ cosθ = (cot θ ) (sin θ ) = 12 ⎛ − 5 ⎞ = − 12 cot θ = cosθ , so sin θ cosθ = (cot θ ) (sin θ ) 4 ⎛ − 3 ⎞ = 4 5 ⎜ 13 ⎟ 13 3 ⎜ 5 ⎟ 5 ⎝ ⎠ tan θ = 1 = 1 = 5 ⎝ ⎠ tan θ = 1 = 1 = − 3 cot θ 12 12 cot θ 4 4 secθ = 1 = 1 = − 13 secθ = 1 = 1 = 5 cosθ 12 12 cosθ 4 35. secθ 5 4 and tan θ < 0 , so θ lies in quadrant II. 37. sin θ = 12 and θ lies in quadrant II. 13 Using the Pythagorean Identities: tan 2 θ = sec2 θ −1 2 Using the Pythagorean Identities: cos2 θ = 1 − sin 2 θ 2 tan 2 θ = ⎛ − 5 ⎞ − 1 = 25 − 1 = 9 cos2 θ = 1 − ⎛ 12 ⎞ = 1 − 144 = 25 ⎜ 4 ⎟ 16 16 ⎜ 13 ⎟ 169 169