We are given:

$\sin(\theta) = \frac{1}{10}$ and $\sec(\theta) < 0.$

Step 1: Analyze the given information

The condition $\sec(\theta) < 0$ implies that $\cos(\theta)$ is negative, because $\sec(\theta) = \frac{1}{\cos(\theta)}$. This means $\theta$ lies either in the second or third quadrant, where $\cos(\theta)$ is negative.

Step 2: Use the Pythagorean identity

We can use the Pythagorean identity to find $\cos(\theta)$. The identity is: $\sin^2(\theta) + \cos^2(\theta) = 1.$ Substituting $\sin(\theta) = \frac{1}{10}$: $\left(\frac{1}{10}\right)^2 + \cos^2(\theta) = 1,$ $\frac{1}{100} + \cos^2(\theta) = 1,$ $\cos^2(\theta) = 1 - \frac{1}{100},$ $\cos^2(\theta) = \frac{99}{100}.$

Step 3: Solve for $\cos(\theta)$

Now, take the square root of both sides: $\cos(\theta) = \pm \frac{\sqrt{99}}{10}.$

Since we know that $\cos(\theta)$ is negative (from the condition $\sec(\theta) < 0$), we have: $\cos(\theta) = -\frac{\sqrt{99}}{10}.$

Step 4: Find $\tan(\theta)$

The tangent function is given by: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.$ Substitute the known values of $\sin(\theta)$ and $\cos(\theta)$: $\tan(\theta) = \frac{\frac{1}{10}}{-\frac{\sqrt{99}}{10}} = -\frac{1}{\sqrt{99}}.$

Thus, the value of $\tan(\theta)$ is: $\tan(\theta) = -\frac{1}{\sqrt{99}}.$

Do you want further details or have any questions? Here are 5 related questions to deepen your understanding:

1. What are the values of sine and cosine in different quadrants?
2. How is the tangent function related to sine and cosine?
3. How would the value of $\tan(\theta)$ change if $\sec(\theta)$ were positive?
4. Can you derive $\sec(\theta)$ from the result of $\cos(\theta)$?
5. What is the exact value of $\sqrt{99}$ to simplify the expression for $\tan(\theta)$?

Tip: When solving trigonometric equations, always analyze the quadrant to determine the sign of the functions.