To find the exact values of $\tan \theta$ and $\sin \theta$ given $\sec \theta = \frac{17}{15}$ and $\cot \theta > 0$, we can follow these steps:

Step 1: Determine the Quadrant

Given that $\sec \theta = \frac{17}{15}$ (which is positive) and $\cot \theta > 0$, we know that both cosine and cotangent are positive. This means that $\theta$ is in the first quadrant because in the first quadrant, all trigonometric functions are positive.

Step 2: Find $\cos \theta$

Since $\sec \theta = \frac{1}{\cos \theta}$, we can find $\cos \theta$:

$\cos \theta = \frac{1}{\sec \theta} = \frac{15}{17}$

Step 3: Find $\sin \theta$

Using the Pythagorean identity:

$\sin^2 \theta + \cos^2 \theta = 1$

Substitute $\cos \theta = \frac{15}{17}$:

$\sin^2 \theta + \left(\frac{15}{17}\right)^2 = 1$

$\sin^2 \theta + \frac{225}{289} = 1$

$\sin^2 \theta = 1 - \frac{225}{289} = \frac{289 - 225}{289} = \frac{64}{289}$

$\sin \theta = \sqrt{\frac{64}{289}} = \frac{8}{17}$

Since $\theta$ is in the first quadrant, $\sin \theta$ is positive:

$\sin \theta = \frac{8}{17}$

Step 4: Find $\tan \theta$

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$\tan \theta = \frac{\frac{8}{17}}{\frac{15}{17}} = \frac{8}{15}$

Conclusion

The exact values are:

$\tan \theta = \frac{8}{15}, \quad \sin \theta = \frac{8}{17}$

Would you like more details on any step or have any questions?

Here are some related questions for further exploration:

1. How do you find the values of other trigonometric functions like $\cot \theta$, $\csc \theta$, and $\cos \theta$?
2. What is the significance of the Pythagorean identity in trigonometry?
3. How can you determine the quadrant of an angle if given different trigonometric conditions?
4. How do the signs of trigonometric functions change across different quadrants?
5. What is the geometric interpretation of these trigonometric values on the unit circle?

Tip: Understanding how to use the Pythagorean identities can greatly simplify solving trigonometric problems involving angles and their respective functions.