Question

Remaining trig functions tanθ= -2√6/3 and cscθ= >0

Answer 1

sin(θ) = 2√22/11

cos(θ) = -√33/11

tan(θ) = -2√6/3 . . . given

sec(θ) = -√33/3

csc(θ) = √22/4

cot(θ) = -√6/4

θ = 128.482°

Explanation:

cot = 1/tan = 1/(-2√6/3) = -3/(2√6) = -√6/4

When the tangent is negative and the cosecant is positive, the angle is in the second quadrant. The cosine and secant are negative there.

sec = -√(1 +tan²) = -√(1 +(-2√6/3)²) = -√(1+24/9) = -√33/3

cos = 1/sec = -3/√33 = -√33/11

sin = tan×cos = (-2√6/3)(-√33/11) = 2√6√33/33 = 2√22/11

csc = 1/sin = 1/(2√22/11) = 11/(2√22) = 11√22/(2·22) = √22/4

Answer 2

• sin(θ) = 2√22/11
• cos(θ) = -√33/11
• tan(θ) = -2√6/3 . . . given
• sec(θ) = -√33/3
• csc(θ) = √22/4
• cot(θ) = -√6/4
• θ = 128.482°

Explanation:

cot = 1/tan = 1/(-2√6/3) = -3/(2√6) = -√6/4

When the tangent is negative and the cosecant is positive, the angle is in the second quadrant. The cosine and secant are negative there.

sec = -√(1 +tan²) = -√(1 +(-2√6/3)²) = -√(1+24/9) = -√33/3

cos = 1/sec = -3/√33 = -√33/11

sin = tan×cos = (-2√6/3)(-√33/11) = 2√6√33/33 = 2√22/11

csc = 1/sin = 1/(2√22/11) = 11/(2√22) = 11√22/(2·22) = √22/4