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Sveta_85 [38]
2 years ago
13

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 .

Mathematics
2 answers:
Nataly_w [17]2 years ago
7 0

Answer:

Part 1) cos(\theta)=-\frac{60}{61}

Part 2) tan(\theta)=\frac{11}{60}  

Part 3) cot(\theta)=\frac{60}{11}

Part 4) csc(\theta)=-\frac{61}{11}

Part 5) sin(\theta)=-\frac{11}{61}  

Step-by-step 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)=\frac{1}{sec(\theta)}

we have

sec(\theta)=-\frac{61}{60} ----> the value must be negative

therefore

cos(\theta)=-\frac{60}{61}

step 2

Find tan(\theta)

we know that

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

we have

sec(\theta)=-\frac{61}{60}

substitute

tan^{2} (\theta)+1=(-\frac{61}{60})^{2}

tan^{2} (\theta)+1=\frac{3,721}{3,600}

tan^{2} (\theta)=\frac{3,721}{3,600}-1

tan^{2} (\theta)=\frac{121}{3,600}

tan(\theta)=\frac{11}{60}

step 3

Find cot(\theta)

we know that

cot(\theta)=\frac{1}{tan(\theta)}

we have

tan(\theta)=\frac{11}{60}

therefore

cot(\theta)=\frac{60}{11}

step 4

Find csc(\theta)

we know that

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

we have

cot(\theta)=\frac{60}{11}

substitute

(\frac{60}{11})^{2}+1=csc^{2} (\theta)

\frac{3,600}{121}+1=csc^{2} (\theta)

\frac{3,721}{121}=csc^{2} (\theta)

square root both sides

csc(\theta)=-\frac{61}{11}

step 5

Find sin(\theta)

we know that

sin(\theta)=\frac{1}{csc(\theta)}

we have

csc(\theta)=-\frac{61}{11}

therefore

sin(\theta)=-\frac{11}{61}

Eva8 [605]2 years ago
7 0

Answer:

Is the answer A?  confused

Step-by-step explanation:

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