Using the sum formula for tangent, express $$ an(a + b) $$ in terms of $$ an(a) $$ and $$ an(b) $$.

Question

Using the sum formula for tangent, express $$ 	an(a + b) $$ in terms of $$ 	an(a) $$ and $$ 	an(b) $$.

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The **sum formula for tangent** allows you to express $$\tan(a + b)$$ in terms of $$\tan(a)$$ and $$\tan(b)$$. The formula is derived from the sine and cosine addition formulas and is given by:

$$
\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}
$$

### Derivation:

1. **Start with the sine and cosine addition formulas:**
   $$
   \sin(a + b) = \sin a \cos b + \cos a \sin b
   $$
   $$
   \cos(a + b) = \cos a \cos b - \sin a \sin b
   $$

2. **Express $$\tan(a + b)$$ using sine and cosine:**
   $$
   \tan(a + b) = \frac{\sin(a + b)}{\cos(a + b)} = \frac{\sin a \cos b + \cos a \sin b}{\cos a \cos b - \sin a \sin b}
   $$

3. **Divide the numerator and the denominator by $$\cos a \cos b$$:**
   $$
   \tan(a + b) = \frac{\frac{\sin a}{\cos a} + \frac{\sin b}{\cos b}}{1 - \frac{\sin a}{\cos a} \cdot \frac{\sin b}{\cos b}} = \frac{\tan a + \tan b}{1 - \tan a \tan b}
   $$

### Final Formula:

$$
\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}
$$

**Example Usage:**

If you know that $$\tan 30^\circ = \frac{1}{\sqrt{3}}$$ and $$\tan 45^\circ = 1$$, you can find $$\tan(30^\circ + 45^\circ) = \tan 75^\circ$$ as follows:

$$
\tan(75^\circ) = \frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}} \cdot 1} = \frac{1 + \sqrt{3}}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = 2 + \sqrt{3}
$$

Thus, $$\tan(75^\circ) = 2 + \sqrt{3}$$.
$$ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} $$

Using the sum formula for tangent, express \( \tan(a + b) \) in terms of \( \tan(a) \) and \( \tan(b) \).

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