Answer:
[tex]tan(\theta)=\sqrt{35}[/tex]
Step-by-step explanation:
we know that
The angle [tex]\theta[/tex] lies in Quadrant III
so
The value of cosine is negative
The value of sine is negative
The value of tangent is positive
we have
[tex]cos(\theta)=-\frac{1}{6}[/tex]
step 1
Find the value of [tex]sin(\theta)[/tex]
Applying trigonometric identity
[tex]sin^2(\theta)+cos^2(\theta)=1[/tex]
substitute the given value
[tex]sin^2(\theta)+(-\frac{1}{6})^2=1[/tex]
[tex]sin^2(\theta)+\frac{1}{36}=1[/tex]
[tex]sin^2(\theta)=1-\frac{1}{36}[/tex]
[tex]sin^2(\theta)=\frac{35}{36}[/tex]
square root both sides
[tex]sin(\theta)=\pm\frac{\sqrt{35}}{6}[/tex]
Remember that the sine is negative (Quadrant III)
so
[tex]sin(\theta)=-\frac{\sqrt{35}}{6}[/tex]
step 2
Find the value of [tex]tan(\theta)[/tex]
we know that
[tex]tan(\theta)=\frac{sin(\theta)}{cos(\theta)}[/tex]
we have
[tex]sin(\theta)=-\frac{\sqrt{35}}{6}[/tex]
[tex]cos(\theta)=-\frac{1}{6}[/tex]
substitute
[tex]tan(\theta)=-\frac{\sqrt{35}}{6}:-\frac{1}{6}=\sqrt{35}[/tex]
