Question

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 .

Answer 1

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

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

Answer 2

Answer:

Is the answer A?  confused

Step-by-step explanation: