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: $tan(\theta) =-\sqrt{3}$

Step-by-step explanation:

In this case we know that:

$sin^2(\theta) = \frac{3}{4}$

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

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

Therefore

$cos^2(\theta)=1-\frac{3}{4}$

$cos^2(\theta)=\frac{1}{4}$

$cos(\theta)=\±\sqrt{\frac{1}{4}}$

$cos(\theta)=\±\frac{1}{2}$

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

Then

$cos(\theta)=-\frac{1}{2}$

$sin(\theta) =\sqrt{\frac{3}{4}}$

$sin(\theta) =\frac{\sqrt{3}}{2}$

Remember that:

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

Finally we have that

$tan(\theta) =\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}$

$tan(\theta) =-\sqrt{3}$