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# Given tan(𝜃)=− square root 7/2 , with sec(𝜃)>0, find sin(𝜃)=?

### 5 Answers

- Wayne DeguManLv 71 month ago
tan𝜃 = -√7/2

sec𝜃 > 0...i.e. 1/cos𝜃 > 0

Hence, cos𝜃 > 0

If tan𝜃 < 0 and cos𝜃 > 0...𝜃 is in Q4

Now, sec²𝜃 = tan²𝜃 + 1

so, sec²𝜃 = (-√7/2)² + 1 => 11/4

Hence, sec𝜃 = ±√11/2

i.e. sec𝜃 = √11/2...as sec𝜃 > 0

Then, cos𝜃 = 2√11/11

Now, tan𝜃 = sin𝜃/cos𝜃

i.e. -√7/2 = sin𝜃/(2√11/11)

Hence, sin𝜃 = -√77/11

Note: sin²𝜃 + cos²𝜃 = 1

checking gives:

(-√77/11)² + (2√11/11)² => 77/121 + 44/121 = 1

:)>

- la consoleLv 71 month ago
sec(θ) > 0

1/cos(θ) > 0

cos(θ) > 0 → given that tan(θ) < 0 → then: sin(θ) < 0

tan(θ) = - √(7/2)

sin(θ)/cos(θ) = - √(7/2)

cos(θ) = - sin(θ)/√(7/2)

cos²(θ) = [- sin(θ)/√(7/2)]²

cos²(θ) = sin²(θ)/(7/2)

cos²(θ) = sin²(θ) * (2/7)

cos²(θ) = (2/7).sin²(θ)

cos²(θ) + sin²(θ) = 1

(2/7).sin²(θ) + sin²(θ) = 1

[(2/7) + (7/7)].sin²(θ) = 1

(9/7).sin²(θ) = 1

sin²(θ) = 7/9

sin(θ) = ± (√7)/3 → recall: sin(θ) < 0

sin(θ) = - (√7)/3

- VamanLv 71 month ago
tan theta=sin theta/cos theta= 7/sqrt(7^2+2^2)= 7/53. cos theta= 2/53. You can guess the other values.

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- JimLv 71 month ago
my answer is "Anonymous" as your post is Anonymous. There is no need for an Anonymous question in this category.