Let's solve the given problem step-by-step:

We are given:

• $\tan \theta = -\frac{3}{8}$
• $\cos \theta < 0$

We need to find the exact values of $\sec \theta$ and $\sin \theta$.

Step 1: Identify the quadrant

Since $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = -\frac{3}{8}$ and $\cos \theta < 0$, we know that the angle $\theta$ must be in the second quadrant, where tangent is negative and cosine is negative.

Step 2: Use the Pythagorean theorem to find the hypotenuse

We can use the relationship between the sides of a right triangle:

$\text{Hypotenuse}^2 = \text{Opposite}^2 + \text{Adjacent}^2$

So, let’s calculate the hypotenuse:

$r^2 = 3^2 + 8^2 = 9 + 64 = 73$ $r = \sqrt{73}$

Step 3: Find $\cos \theta$ and $\sin \theta$

In the second quadrant:

• $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{-8}{\sqrt{73}}$
• $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{\sqrt{73}}$

Step 4: Find $\sec \theta$

$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{-8}{\sqrt{73}}} = \frac{-\sqrt{73}}{8}$

Thus: $\sec \theta = \frac{-\sqrt{73}}{8}$ $\sin \theta = \frac{3}{\sqrt{73}}$

These are the exact values for $\sec \theta$ and $\sin \theta$.

Do you need any further details on this solution? Here are some related questions:

1. How do we use the Pythagorean identity to solve trigonometric problems?
2. Why is cosine negative in the second quadrant?
3. How can we rationalize a trigonometric expression involving a square root in the denominator?
4. What are the characteristics of angles in different quadrants in the unit circle?
5. Can we use this approach to find other trigonometric functions for $\theta$?

Tip: Always remember to check the quadrant when dealing with trigonometric problems, as it determines the signs of the trigonometric values.