It looks like you have two trigonometric questions in the image:

Question 1:

Given:

• $\tan(\theta) = \frac{7}{6}$
• Quadrant: III

You need to find $\sec(\theta)$.

Steps:

1. We know that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$, so $\sin(\theta) = 7k$ and $\cos(\theta) = 6k$ for some constant $k$.
2. Using the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we substitute: $(7k)^2 + (6k)^2 = 1$ $49k^2 + 36k^2 = 1$ $85k^2 = 1 \implies k^2 = \frac{1}{85} \implies k = \frac{1}{\sqrt{85}}$
3. Now, find $\sec(\theta) = \frac{1}{\cos(\theta)}$. Since $\cos(\theta) = 6k$, we have: $\sec(\theta) = \frac{1}{6k} = \frac{1}{6 \times \frac{1}{\sqrt{85}}} = \frac{\sqrt{85}}{6}$
4. In the third quadrant, $\sec(\theta)$ is negative because cosine is negative in this quadrant. Thus, the answer is: $\sec(\theta) = -\frac{\sqrt{85}}{6}$

Question 2:

Given:

• $\cos(\theta) = \frac{3}{4}$
• Quadrant: I

You need to find $\csc(\theta)$.

Steps:

1. Use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$. Since $\cos(\theta) = \frac{3}{4}$, we can find $\sin(\theta)$: $\sin^2(\theta) + \left( \frac{3}{4} \right)^2 = 1$ $\sin^2(\theta) + \frac{9}{16} = 1$ $\sin^2(\theta) = 1 - \frac{9}{16} = \frac{7}{16}$ $\sin(\theta) = \frac{\sqrt{7}}{4}$
2. Now, find $\csc(\theta) = \frac{1}{\sin(\theta)}$: $\csc(\theta) = \frac{1}{\frac{\sqrt{7}}{4}} = \frac{4}{\sqrt{7}} = \frac{4\sqrt{7}}{7}$

Thus, the answers are:

• $\sec(\theta) = -\frac{\sqrt{85}}{6}$ (from the first question)
• $\csc(\theta) = \frac{4\sqrt{7}}{7}$ (from the second question)

Would you like more details or explanations on any step? Let me know if you have any questions!

Here are 5 related questions to explore further:

1. How do you derive trigonometric identities using Pythagorean identities?
2. How does the sign of trigonometric functions change in different quadrants?
3. What are other ways to express secant and cosecant in terms of sine and cosine?
4. How do you find trigonometric values when given a point on the unit circle?
5. How do you handle negative angles in trigonometric functions?

Tip: Always check which quadrant the angle lies in before determining the sign of the trigonometric functions!