Answer:
Part 1) [tex]cos(\theta)=-\frac{60}{61}[/tex]
Part 2) [tex]tan(\theta)=\frac{11}{60}[/tex]
Part 3) [tex]cot(\theta)=\frac{60}{11}[/tex]
Part 4) [tex]csc(\theta)=-\frac{61}{11}[/tex]
Part 5) [tex]sin(\theta)=-\frac{11}{61}[/tex]
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 [tex]cos(\theta)[/tex]
we know that
[tex]cos(\theta)=\frac{1}{sec(\theta)}[/tex]
we have
[tex]sec(\theta)=-\frac{61}{60}[/tex] ----> the value must be negative
therefore
[tex]cos(\theta)=-\frac{60}{61}[/tex]
step 2
Find [tex]tan(\theta)[/tex]
we know that
[tex]tan^{2} (\theta)+1=sec^{2} (\theta)[/tex]
we have
[tex]sec(\theta)=-\frac{61}{60}[/tex]
substitute
[tex]tan^{2} (\theta)+1=(-\frac{61}{60})^{2}[/tex]
[tex]tan^{2} (\theta)+1=\frac{3,721}{3,600}[/tex]
[tex]tan^{2} (\theta)=\frac{3,721}{3,600}-1[/tex]
[tex]tan^{2} (\theta)=\frac{121}{3,600}[/tex]
[tex]tan(\theta)=\frac{11}{60}[/tex]
step 3
Find [tex]cot(\theta)[/tex]
we know that
[tex]cot(\theta)=\frac{1}{tan(\theta)}[/tex]
we have
[tex]tan(\theta)=\frac{11}{60}[/tex]
therefore
[tex]cot(\theta)=\frac{60}{11}[/tex]
step 4
Find [tex]csc(\theta)[/tex]
we know that
[tex]cot^{2} (\theta)+1=csc^{2} (\theta)[/tex]
we have
[tex]cot(\theta)=\frac{60}{11}[/tex]
substitute
[tex](\frac{60}{11})^{2}+1=csc^{2} (\theta)[/tex]
[tex]\frac{3,600}{121}+1=csc^{2} (\theta)[/tex]
[tex]\frac{3,721}{121}=csc^{2} (\theta)[/tex]
square root both sides
[tex]csc(\theta)=-\frac{61}{11}[/tex]
step 5
Find [tex]sin(\theta)[/tex]
we know that
[tex]sin(\theta)=\frac{1}{csc(\theta)}[/tex]
we have
[tex]csc(\theta)=-\frac{61}{11}[/tex]
therefore
[tex]sin(\theta)=-\frac{11}{61}[/tex]
