In this problem, we're given some trigonometric information to determine the remaining trigonometric values for angle $\theta$. Here’s what we know:

• $\cos \theta = -\frac{1}{4}$
• $\tan \theta > 0$ (which indicates $\theta$ is in the third quadrant where both sine and cosine are negative, but tangent is positive because it's the ratio of sine to cosine)

Step-by-step Solution:

1. Using $\cos \theta = -\frac{1}{4}$ and the identity $\sin^2 \theta + \cos^2 \theta = 1$, we can find $\sin \theta$.

   $\sin^2 \theta + \left( -\frac{1}{4} \right)^2 = 1$ $\sin^2 \theta + \frac{1}{16} = 1$ $\sin^2 \theta = 1 - \frac{1}{16} = \frac{15}{16}$ $\sin \theta = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}$

   (We take the negative square root because in the third quadrant, sine is negative.)

2. Now, using $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

   $\tan \theta = \frac{-\frac{\sqrt{15}}{4}}{-\frac{1}{4}} = \sqrt{15}$

3. For the reciprocal identities:

   • $\csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{\sqrt{15}}{4}} = -\frac{4}{\sqrt{15}} = -\frac{4\sqrt{15}}{15}$
   • $\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{1}{4}} = -4$
   • $\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15}$

Final Values:

• $\sin \theta = -\frac{\sqrt{15}}{4}$
• $\cos \theta = -\frac{1}{4}$
• $\tan \theta = \sqrt{15}$
• $\csc \theta = -\frac{4\sqrt{15}}{15}$
• $\sec \theta = -4$
• $\cot \theta = \frac{\sqrt{15}}{15}$

Would you like any further explanation or details on these calculations?

Here are 5 related questions to explore further:

1. How do you determine the quadrant of an angle given conditions on the signs of sine, cosine, and tangent?
2. What are the reciprocal trigonometric identities?
3. How is the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ used to find unknown trigonometric values?
4. What are the properties of trigonometric functions in different quadrants?
5. How can you simplify expressions involving square roots, like $\frac{4}{\sqrt{15}}$?

Tip: When solving trigonometric problems, always carefully consider the quadrant to determine the correct signs for the trigonometric functions!