Certainly! The equation provided involves simplifying $\tan \left( \frac{\pi}{12} \right)$ by breaking it into a difference of two angles:

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$\tan \left( \frac{\pi}{12} \right) = \tan \left( \frac{4\pi}{12} - \frac{3\pi}{12} \right)$

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Here's a step-by-step breakdown of why this makes sense:

1. Identify the Two Angles

• Notice that $\frac{\pi}{12}$ is being split into a difference of two angles:
  $\frac{\pi}{12} = \frac{4\pi}{12} - \frac{3\pi}{12}$
• Simplifying each term:
  • $\frac{4\pi}{12} = \frac{\pi}{3}$
  • $\frac{3\pi}{12} = \frac{\pi}{4}$

Thus: $\tan \left( \frac{\pi}{12} \right) = \tan \left( \frac{\pi}{3} - \frac{\pi}{4} \right)$

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2. Use the Tangent Difference Formula

The tangent difference formula is:
$\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$
Here:

• $A = \frac{\pi}{3}$
• $B = \frac{\pi}{4}$

We know from trigonometric values:

• $\tan \left( \frac{\pi}{3} \right) = \sqrt{3}$
• $\tan \left( \frac{\pi}{4} \right) = 1$

Substitute these into the formula: $\tan \left( \frac{\pi}{3} - \frac{\pi}{4} \right) = \frac{\tan \left( \frac{\pi}{3} \right) - \tan \left( \frac{\pi}{4} \right)}{1 + \tan \left( \frac{\pi}{3} \right) \tan \left( \frac{\pi}{4} \right)}$

$\tan \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} - 1}{1 + \sqrt{3} \cdot 1}$

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3. Simplify the Expression

• Numerator: $\sqrt{3} - 1$
• Denominator: $1 + \sqrt{3}$

Thus: $\tan \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} - 1}{1 + \sqrt{3}}$

To further simplify, you can rationalize the denominator by multiplying numerator and denominator by $1 - \sqrt{3}$.

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Final Notes

The key idea here is breaking $\frac{\pi}{12}$ into a difference of two angles where tangent values are well-known. Then, applying the tangent difference formula allows for simplification.

Let me know if you'd like the rationalized result or a deeper explanation!

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Related Questions:

1. What is the tangent addition formula, and how is it derived?
2. How do you simplify $\tan (A + B)$ and $\tan (A - B)$ using known values?
3. What are the exact values of tangent for common angles like $\frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{6}$, etc.?
4. How do you rationalize expressions with square roots in the denominator?
5. Can you derive the tangent difference formula step by step?

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Tip:

Always remember the exact values for sine, cosine, and tangent for special angles (like $\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$). They are fundamental for solving trigonometric problems!