To find the exact value of the expression, $\tan \left( \frac{\pi}{6} + \frac{3\pi}{4} \right)$, we will use the tangent addition formula:

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

Step 1: Identify the values

• $a = \frac{\pi}{6}$
• $b = \frac{3\pi}{4}$

Step 2: Find $\tan a$ and $\tan b$

• $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$
• $\tan \frac{3\pi}{4} = -1$

Step 3: Apply the tangent addition formula

Substitute the values into the formula:

$\tan \left( \frac{\pi}{6} + \frac{3\pi}{4} \right) = \frac{\frac{1}{\sqrt{3}} + (-1)}{1 - \left( \frac{1}{\sqrt{3}} \cdot -1 \right)}$

Simplify each part of the fraction:

1. Numerator: $\frac{1}{\sqrt{3}} - 1 = \frac{1 - \sqrt{3}}{\sqrt{3}}$
2. Denominator: $1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}$

So we have:

$\tan \left( \frac{\pi}{6} + \frac{3\pi}{4} \right) = \frac{\frac{1 - \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}$

Step 4: Simplify the expression

Since both the numerator and the denominator are divided by $\sqrt{3}$, they cancel out:

$\tan \left( \frac{\pi}{6} + \frac{3\pi}{4} \right) = \frac{1 - \sqrt{3}}{\sqrt{3} + 1}$

Step 5: Rationalize the denominator

Multiply the numerator and denominator by $\sqrt{3} - 1$ to eliminate the square root in the denominator:

$= \frac{(1 - \sqrt{3})(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)}$

Expanding both the numerator and the denominator:

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

So we have:

$\tan \left( \frac{\pi}{6} + \frac{3\pi}{4} \right) = \frac{-4 + 2\sqrt{3}}{2} = -2 + \sqrt{3}$

Final Answer:

$\tan \left( \frac{\pi}{6} + \frac{3\pi}{4} \right) = -2 + \sqrt{3}$

Would you like further details on this solution, or do you have any questions?

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Here are 5 related questions to explore further:

1. What is the tangent addition formula and why is it used in trigonometry?
2. How can we rationalize denominators with complex expressions?
3. How would the solution change if we used degrees instead of radians?
4. How do we simplify radicals in trigonometric expressions?
5. What are other methods for calculating tangent values of special angle sums?

Tip: For complex trigonometric expressions, breaking down each step helps simplify and minimize errors.