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Suppose θ is an angle in the standard position whose terminal side is in Quadrant III and sec θ=61/60. Find the exact values of the five remaining trigonometric functions of θ .

Suppose θ is an angle in the standard position whose terminal side is in Quadrant-example-1

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Answer:

Is the answer A? confused

Explanation:

User DerKuchen
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3 votes

Answer:

Part 1)
cos(\theta)=-(60)/(61)

Part 2)
tan(\theta)=(11)/(60)

Part 3)
cot(\theta)=(60)/(11)

Part 4)
csc(\theta)=-(61)/(11)

Part 5)
sin(\theta)=-(11)/(61)

Explanation:

we know that

If angle theta lie on Quadrant III

then

The function sine is negative

The function cosine is negative

The function tangent is positive

The function cotangent is positive

The function cosecant is negative

The function secant is negative

step 1

Find
cos(\theta)

we know that


cos(\theta)=(1)/(sec(\theta))

we have


sec(\theta)=-(61)/(60) ----> the value must be negative

therefore


cos(\theta)=-(60)/(61)

step 2

Find
tan(\theta)

we know that


tan^(2) (\theta)+1=sec^(2) (\theta)

we have


sec(\theta)=-(61)/(60)

substitute


tan^(2) (\theta)+1=(-(61)/(60))^(2)


tan^(2) (\theta)+1=(3,721)/(3,600)


tan^(2) (\theta)=(3,721)/(3,600)-1


tan^(2) (\theta)=(121)/(3,600)


tan(\theta)=(11)/(60)

step 3

Find
cot(\theta)

we know that


cot(\theta)=(1)/(tan(\theta))

we have


tan(\theta)=(11)/(60)

therefore


cot(\theta)=(60)/(11)

step 4

Find
csc(\theta)

we know that


cot^(2) (\theta)+1=csc^(2) (\theta)

we have


cot(\theta)=(60)/(11)

substitute


((60)/(11))^(2)+1=csc^(2) (\theta)


(3,600)/(121)+1=csc^(2) (\theta)


(3,721)/(121)=csc^(2) (\theta)

square root both sides


csc(\theta)=-(61)/(11)

step 5

Find
sin(\theta)

we know that


sin(\theta)=(1)/(csc(\theta))

we have


csc(\theta)=-(61)/(11)

therefore


sin(\theta)=-(11)/(61)

User Elsni
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