To compute \(\tan(-\pi/12)\), we can use the **angle addition formula** for tangent:

\[
\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}.
\]

Here, we break \(-\pi/12\) into two angles with known tangent values, such as:

\[
-\frac{\pi}{12} = -\frac{\pi}{4} + \frac{\pi}{6}.
\]

### Step-by-Step Solution:
1. **Use the tangent addition formula**:
   \[
   \tan(-\frac{\pi}{12}) = \tan(-\frac{\pi}{4} + \frac{\pi}{6}) = \frac{\tan(-\frac{\pi}{4}) + \tan(\frac{\pi}{6})}{1 - \tan(-\frac{\pi}{4})\tan(\frac{\pi}{6})}.
   \]

2. **Recall values of tangent**:
   \[
   \tan(-\frac{\pi}{4}) = -1, \quad \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}.
   \]

3. **Substitute these values**:
   \[
   \tan(-\frac{\pi}{12}) = \frac{-1 + \frac{1}{\sqrt{3}}}{1 - (-1) \cdot \frac{1}{\sqrt{3}}}.
   \]

4. **Simplify the numerator and denominator**:
   \[
   \text{Numerator: } -1 + \frac{1}{\sqrt{3}} = \frac{-\sqrt{3} + 1}{\sqrt{3}}.
   \]
   \[
   \text{Denominator: } 1 - (-1 \cdot \frac{1}{\sqrt{3}}) = 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}.
   \]

5. **Divide the two fractions**:
   \[
   \tan(-\frac{\pi}{12}) = \frac{\frac{-\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{-\sqrt{3} + 1}{\sqrt{3} + 1}.
   \]

### Final Answer:
\[
\tan(-\frac{\pi}{12}) = \frac{-\sqrt{3} + 1}{\sqrt{3} + 1}.
\]

---

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### Related Questions:
1. How is the tangent addition formula derived?
2. What are the exact values of \(\tan(\pi/6)\) and \(\tan(\pi/4)\)?
3. Can you calculate \(\tan(\pi/12)\) and compare with \(\tan(-\pi/12)\)?
4. What is the geometric interpretation of \(-\pi/12\) on the unit circle?
5. How does the periodicity of the tangent function affect its values?

### Tip:
The tangent function is periodic with a period of \(\pi\), so \(\tan(\theta + \pi) = \tan(\theta)\). Use this property to simplify expressions with angles outside the standard range!

Discover how to calculate tan(-π/12) using the angle addition formula for tangent. This step-by-step solution demonstrates key trigonometric concepts, including exact values of tan(π/4) and tan(π/6), and simplifies the result to (-√3 + 1)/(√3 + 1). Ideal for Grades 10-12 students exploring trigonometric identities.

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