Question

if theta in quadrant iI and sin^2theta = 3/4, evaluate tan theta in your final answer, include an explanation of the solution.

Answer 1

Answer:
`tan(\theta) =-√(3)`

Explanation:

In this case we know that:

`sin^2(\theta) = (3)/(4)`

To find the value of
`cos(\theta)` we use the following trigonometric identity

`cos^2(\theta)=1-sin^2(\theta)`

Therefore

`cos^2(\theta)=1-(3)/(4)`

`cos^2(\theta)=(1)/(4)`

`cos(\theta)=\±\sqrt{(1)/(4)}`

`cos(\theta)=\±(1)/(2)`

In the second quadrant
`cos(\theta)<0` and
`sin(\theta)>0`

Then

`cos(\theta)=-(1)/(2)`

`sin(\theta) =\sqrt{(3)/(4)}`

`sin(\theta) =(√(3))/(2)`

Remember that:

`tan(\theta) =(sin(\theta))/(cos(\theta))`

Finally we have that

`tan(\theta) =((√(3))/(2))/(-(1)/(2))`

`tan(\theta) =-√(3)`