Question

Find cot θ if csc θ = sqrt5/2 and tan θ > 0.

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Answer 1

@jdoe0001 @DebbieG

Answer 2

\[ csc\theta = \sqrt{\frac{5}{2}} \quad \text{?}\]

Answer 3

@pitamar just sqrt on the 5

Answer 4

Well, as with the last one, my preferred method would be to use the appropriate pythagorean identity. In this case, it would be the one that relates cotangent and cosecant:

\(\large \cot^2 x + 1=\csc^2 x\)

Use it to solve for \(\large \cot x \), which will get you to within a \(\pm\). Then use the fact that you can determine the quadrant to get the correct sign. Since cosecant >0 and tangent >0, you know that the angle is in which quadrant?

Answer 5

@pitamar can you give me the answer and how you found it?

Answer 6

I'm not the one who answered. did you mean @DebbieG ?

Answer 7

I know, but i was just wondering if you knew the answer

Answer 8

I'm typing it

Answer 9

Okay thank you

Answer 10

ffs.. one small fix... forgot +- somewhere..
\[
csc(\theta) = \frac{ \sqrt{5}}{2} = \frac{ \text{hypotenuse }}{ \text{side} } \\
sin (\theta) = \frac {2}{ \sqrt{5}} = \frac{ \text{side} }{ \text{hypotenuse}} \\
cos(\theta) = \pm\sqrt{ 1 - sin^2(\theta) } \\
cos(\theta) = \pm\sqrt{ 1 - (\frac{2}{\sqrt{5}})^2 } = \pm\sqrt{ 1 - \frac{4}{5} } = \pm\sqrt{ \frac{1}{5} } = \pm\frac{1}{\sqrt{5}} \\
[tan(\theta) > 0 \quad and \quad sin(\theta) > 0] \quad \to \quad cos(\theta) > 0 \\
cos(\theta) = \frac{1}{ \sqrt{5} } \\
cot(\theta) = \frac{cos(\theta)}{sin(\theta)} = cos(\theta) \cdot \frac{1}{sin(\theta)} =
\frac{1}{ \sqrt{5} } \cdot \frac{ \sqrt{5} }{2} = \frac{1}{2}
\]

Answer 11

Thank you so much