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

Question

asked Jan 8, 2020 67.3k views

1 Answer

Lonami · Answer 1 · 2020-01-12T17:39:53+0000

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

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

$tan(\theta) =-√(3)$