The question given is asking to solve for trigonometric values, given the information:

• $\sin \theta = -\frac{3}{4}$
• $\tan \theta > 0$

This implies that $\theta$ lies in the third quadrant, where both sine and cosine are negative, and tangent is positive.

Let’s compute all the required values:

1. Using the Pythagorean Identity:

$\sin^2\theta + \cos^2\theta = 1$ Substitute $\sin\theta = -\frac{3}{4}$: $\left(-\frac{3}{4}\right)^2 + \cos^2\theta = 1$ $\frac{9}{16} + \cos^2\theta = 1$ $\cos^2\theta = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16}$ $\cos\theta = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4} \quad \text{(negative in the third quadrant)}.$

2. Compute $\tan\theta$:

$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-\frac{3}{4}}{-\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}} = \frac{3\sqrt{7}}{7}.$

3. Compute $\cot\theta$:

$\cot\theta = \frac{1}{\tan\theta} = \frac{\sqrt{7}}{3}.$

4. Compute $\sec\theta$:

$\sec\theta = \frac{1}{\cos\theta} = \frac{1}{-\frac{\sqrt{7}}{4}} = -\frac{4}{\sqrt{7}} = -\frac{4\sqrt{7}}{7}.$

5. Compute $\csc\theta$:

$\csc\theta = \frac{1}{\sin\theta} = \frac{1}{-\frac{3}{4}} = -\frac{4}{3}.$

Summary of Results:

• $\sin\theta = -\frac{3}{4}$
• $\cos\theta = -\frac{\sqrt{7}}{4}$
• $\tan\theta = \frac{3\sqrt{7}}{7}$
• $\cot\theta = \frac{\sqrt{7}}{3}$
• $\sec\theta = -\frac{4\sqrt{7}}{7}$
• $\csc\theta = -\frac{4}{3}$

Would you like further clarification or detailed steps for any part of the solution?

Related Questions:

1. What are the quadrant-specific signs of trigonometric functions?
2. How do you use the Pythagorean identity for solving trigonometric values?
3. Why is $\tan\theta$ positive while $\sin\theta$ and $\cos\theta$ are negative?
4. How can rationalizing fractions like $\frac{3}{\sqrt{7}}$ be simplified for answers?
5. What is the importance of the unit circle in solving trigonometric equations?

Tip: Always double-check the quadrant to assign the correct signs to $\sin\theta$, $\cos\theta$, and other values!