To find the exact value of $\tan \left( \frac{\pi}{12} \right)$, we can use the tangent difference identity:

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$

Let's rewrite $\frac{\pi}{12}$ as a difference of two angles where we know the tangent values. We can use:

$\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{3}$

Now we apply the identity:

$\tan \left( \frac{\pi}{12} \right) = \tan \left( \frac{\pi}{4} - \frac{\pi}{3} \right) = \frac{\tan \frac{\pi}{4} - \tan \frac{\pi}{3}}{1 + \tan \frac{\pi}{4} \cdot \tan \frac{\pi}{3}}$

Substitute the values:

• $\tan \frac{\pi}{4} = 1$
• $\tan \frac{\pi}{3} = \sqrt{3}$

So,

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

To simplify, we can rationalize the denominator:

$\tan \left( \frac{\pi}{12} \right) = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \cdot \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{(1 - \sqrt{3})^2}{(1 + \sqrt{3})(1 - \sqrt{3})}$

Expanding both the numerator and the denominator:

1. Numerator: $(1 - \sqrt{3})^2 = 1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3}$
2. Denominator: $(1 + \sqrt{3})(1 - \sqrt{3}) = 1 - 3 = -2$

Thus,

$\tan \left( \frac{\pi}{12} \right) = \frac{4 - 2\sqrt{3}}{-2} = -2 + \sqrt{3}$

The exact value of $\tan \left( \frac{\pi}{12} \right)$ is:

$\tan \left( \frac{\pi}{12} \right) = \sqrt{3} - 2$

Would you like further clarification or have any additional questions?

Here are five related questions that might help expand your understanding:

1. How would you find $\sin \left( \frac{\pi}{12} \right)$ using a similar approach?
2. Can the sum identity for cosine be used to verify the result for $\tan \left( \frac{\pi}{12} \right)$?
3. How does the tangent difference identity differ from the sum identity?
4. What is the exact value of $\tan \left( \frac{\pi}{6} \right)$ using known values?
5. Why do we rationalize the denominator when working with trigonometric values?

Tip: Familiarizing yourself with trigonometric identities, like the sum and difference identities, can simplify solving these kinds of exact value problems significantly.