Find sin θ if cot θ = - 2 and cos θ < 0.

Solution:

Given that cot θ = - 2 and cos θ < 0

Clearly cot θ and cos θ both are negative.

⇒ θ is an angle from II quadrant

From OA² = OB² + AB²

OA² = (-2)² + 1²

OA² = 5 ⇒ OA = √5

cot θ = Adjacent side / Opposite = -2 / 1

sin θ = Opposite / Hypotenuse = 1 / √5

Aliter:

cos θ < 0 and cot θ = - 2, tells us that sine quantity must be positive and the quadrant will be the 2nd quadrant.

We will start with the trigonometric identity: 1 + cot² (θ) = csc² (θ)

Subsitute cot² (θ) = (-2)² 

1 + (-2)² = csc² (θ)

5 = csc² (θ)

Substitute csc² (θ) = 1 / sin² (θ)

5 = 1 / sin² (θ)

sin² (θ) = 1 / 5

sin (θ) = ±1 / √5

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Find sin θ if cot θ = - 2 and cos θ < 0.

Summary: 

If cot θ = - 2 and cos θ < 0, sin θ = 1 / √5.