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Please Help me and show your work

Please Help me and show your work-example-1
User P Basak
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Given cotθ = -3 and cosθ < 0:

sinθ = -√(1 - cos²θ)

cosθ < 0

tanθ = -1/3

cotθ = -3

secθ = -1/cosθ

cscθ = -1/sinθ

We are given that cot(θ) = -3 and cos(θ) < 0. Let's evaluate the six trigonometric functions based on this information.

Sine (sinθ):

We know that

cot(θ) = 1/tan(θ), and tan(θ) = 1/cot(θ).

Therefore, tan(θ) = -1/3.

Using the Pythagorean identity sin²θ + cos²θ = 1,

We find sinθ as √(1 - cos²θ).

Given that cos(θ) < 0, we take the negative square root: sinθ = -√(1 - cos²θ).

Cosine (cosθ):

We are given cos(θ) < 0.

Tangent (tanθ):

tan(θ) = -1/3.

Cotangent (cotθ):

cot(θ) = -3 (given).

Secant (secθ):

sec(θ) is the reciprocal of cos(θ). Since cos(θ) < 0, sec(θ) = -1/cos(θ).

Cosecant (cscθ):

csc(θ) is the reciprocal of sin(θ). Therefore, cscθ = -1/sinθ.

User Jtomasrl
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