Question

The angle 0 lies in Quadrant III cos 0 =1/6 What is tan ?

Answer 1

Answer:

$tan(\theta)=\sqrt{35}$

Step-by-step explanation:

we know that

The angle $\theta$ 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

$cos(\theta)=-\frac{1}{6}$

step 1

Find the value of $sin(\theta)$

Applying trigonometric identity

$sin^2(\theta)+cos^2(\theta)=1$

substitute the given value

$sin^2(\theta)+(-\frac{1}{6})^2=1$

$sin^2(\theta)+\frac{1}{36}=1$

$sin^2(\theta)=1-\frac{1}{36}$

$sin^2(\theta)=\frac{35}{36}$

square root both sides

$sin(\theta)=\pm\frac{\sqrt{35}}{6}$

Remember that the sine is negative (Quadrant III)

so

$sin(\theta)=-\frac{\sqrt{35}}{6}$

step 2

Find the value of $tan(\theta)$

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=-\frac{\sqrt{35}}{6}$

$cos(\theta)=-\frac{1}{6}$

substitute

$tan(\theta)=-\frac{\sqrt{35}}{6}:-\frac{1}{6}=\sqrt{35}$