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Mathematics 19 Online
OpenStudy (anonymous):

How do you find the cot θ if csc θ = square root of five divided by two and tan θ > 0.

OpenStudy (freckles):

do you want to use Pythagorean identities or do you want to use a drawing.

OpenStudy (anonymous):

Pythagorean identities please

OpenStudy (freckles):

well then we know the following: \[\sin^2(\theta)+\cos^2(\theta)=1 \\ 1+\cot^2(\theta)=\csc^2(\theta) \\ \tan^2(\theta)+1=\sec^2(\theta)\]

OpenStudy (freckles):

which equation I just listed do you think would be most helpful here

OpenStudy (anonymous):

the second

OpenStudy (freckles):

exactly now let's keep in mind this: if tan is pos then cot is pos if tan is neg then cot is neg here we are given that tan>0 which means tan is pos

OpenStudy (freckles):

so we already know what sign the answer will have

OpenStudy (freckles):

now we need to replace csc(theta) with sqrt(5)/2

OpenStudy (freckles):

\[1+\cot^2(\theta)=(\frac{\sqrt{5}}{2})^2 \]

OpenStudy (freckles):

can you simplify the right hand side before we continue

OpenStudy (anonymous):

yah

OpenStudy (freckles):

so once you do that you will solve that equation like you would if you were solving the following for x: \[1+x^2=(\frac{\sqrt{5}}{2})^2 \\ x^2=(\frac{\sqrt{5}}{2})^2-1 \] you would then take square root of both sides and decide which sign to use

OpenStudy (anonymous):

so the right side would be 5/4

OpenStudy (freckles):

yah so you have \[1+\cot^2(\theta)=\frac{5}{4}\]

OpenStudy (anonymous):

then how would you continue?

OpenStudy (freckles):

just like I did above

OpenStudy (freckles):

subtract 1 on both sides

OpenStudy (freckles):

then take the square root of both sides and decide which sign to use

OpenStudy (anonymous):

ok one second

OpenStudy (anonymous):

\[\cot \theta=\sqrt{5/4-1}\]

OpenStudy (freckles):

right you chose the right sign since cot is positive

OpenStudy (freckles):

now simplify your answer

OpenStudy (anonymous):

it would be \[1/2\]

OpenStudy (freckles):

yep yep

OpenStudy (anonymous):

Thank you XD

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