Let’s solve the problem step by step.

The given angle is $\theta = \frac{7\pi}{6}$. This angle lies in the third quadrant because $\frac{7\pi}{6}$ is slightly more than $\pi$ ($\pi = \frac{6\pi}{6}$). The reference angle is:

$\text{Reference angle} = \theta - \pi = \frac{7\pi}{6} - \pi = \frac{\pi}{6}.$

The trigonometric functions in the third quadrant are as follows:

• Sine and cosecant ($\sin \theta, \csc \theta$) are negative.
• Cosine and secant ($\cos \theta, \sec \theta$) are negative.
• Tangent and cotangent ($\tan \theta, \cot \theta$) are positive.

The exact values for $\sin$, $\cos$, and $\tan$ of $\frac{\pi}{6}$ (from reference angles) are: $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}.$

Using these values:

1. $\sec(\theta)$: $\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}.$

2. $\csc(\theta)$: $\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{-\frac{1}{2}} = -2.$

3. $\tan(\theta)$: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.$

4. $\cot(\theta)$: $\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{\frac{\sqrt{3}}{3}} = \sqrt{3}.$

Final answers:

$\sec(\theta) = -\frac{2\sqrt{3}}{3}, \quad \csc(\theta) = -2, \quad \tan(\theta) = \frac{\sqrt{3}}{3}, \quad \cot(\theta) = \sqrt{3}.$

Let me know if you'd like more details or have questions.

Related Questions:

1. What is the reference angle for $\theta = \frac{5\pi}{4}$?
2. How do you determine the quadrant of a given angle?
3. What are the signs of trigonometric functions in all quadrants?
4. How is the secant function related to cosine?
5. Why does the tangent of $\frac{\pi}{6}$ simplify to $\frac{\sqrt{3}}{3}$?

Tip: Always start by finding the reference angle and determining the quadrant when solving trigonometric problems!