We are tasked with finding $\tan \theta$ given that:

$\sin \theta = -\frac{3}{4} \quad \text{and} \quad \theta \text{ is in Quadrant II}.$

Step-by-Step Solution:

1. Use the Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$ Substitute $\sin \theta = -\frac{3}{4}$: $\left(-\frac{3}{4}\right)^2 + \cos^2 \theta = 1$ $\frac{9}{16} + \cos^2 \theta = 1$ Solving for $\cos^2 \theta$: $\cos^2 \theta = 1 - \frac{9}{16} = \frac{7}{16}$ Thus: $\cos \theta = \pm \frac{\sqrt{7}}{4}$

2. Determine the sign of $\cos \theta$: Since $\theta$ is in Quadrant II, cosine is negative in this quadrant. Therefore: $\cos \theta = -\frac{\sqrt{7}}{4}$

3. Find $\tan \theta$: The tangent function is defined as: $\tan \theta = \frac{\sin \theta}{\cos \theta}$ Substituting the values: $\tan \theta = \frac{-\frac{3}{4}}{-\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}}$ To rationalize the denominator: $\tan \theta = \frac{3}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{3 \sqrt{7}}{7}$

Thus, the value of $\tan \theta$ is:

$\tan \theta = \frac{3 \sqrt{7}}{7}$

Would you like further details or have any questions?

Here are 5 related questions:

1. How can we find $\cot \theta$ given the value of $\tan \theta$?
2. What is the value of $\sec \theta$ for the given problem?
3. How do we determine the quadrant of an angle based on the signs of sine and cosine?
4. Can you explain why cosine is negative in Quadrant II?
5. How can we use the Pythagorean identity to find $\cos \theta$ for any trigonometric function?

Tip: Always remember that the signs of trigonometric functions depend on the quadrant in which the angle lies.