To find the exact value of \(\tan\left(\frac{5\pi}{6} + \frac{3\pi}{4}\right)\), we can use the tangent addition formula:

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

Let's apply this to the given problem where \( A = \frac{5\pi}{6} \) and \( B = \frac{3\pi}{4} \).

### Step 1: Find \(\tan\left(\frac{5\pi}{6}\right)\)
The angle \(\frac{5\pi}{6}\) is in the second quadrant, where the tangent function is negative.

\[
\tan\left(\frac{5\pi}{6}\right) = \tan\left(\pi - \frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}
\]

### Step 2: Find \(\tan\left(\frac{3\pi}{4}\right)\)
The angle \(\frac{3\pi}{4}\) is also in the second quadrant, where the tangent function is negative.

\[
\tan\left(\frac{3\pi}{4}\right) = \tan\left(\pi - \frac{\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) = -1
\]

### Step 3: Apply the Tangent Addition Formula
Substitute the values into the formula:

\[
\tan\left(\frac{5\pi}{6} + \frac{3\pi}{4}\right) = \frac{\tan\left(\frac{5\pi}{6}\right) + \tan\left(\frac{3\pi}{4}\right)}{1 - \tan\left(\frac{5\pi}{6}\right) \tan\left(\frac{3\pi}{4}\right)}
\]

\[
= \frac{-\frac{1}{\sqrt{3}} + (-1)}{1 - \left(-\frac{1}{\sqrt{3}}\right)(-1)}
\]

\[
= \frac{-\frac{1}{\sqrt{3}} - 1}{1 - \frac{1}{\sqrt{3}}}
\]

### Step 4: Simplify the Expression
To simplify the expression, multiply both the numerator and denominator by \(\sqrt{3}\) to get rid of the square roots:

\[
= \frac{-\frac{\sqrt{3}}{3} - \sqrt{3}}{\sqrt{3} - 1}
\]

Factor out \(\sqrt{3}\) from the numerator:

\[
= \frac{-\sqrt{3}\left(\frac{1}{3} + 1\right)}{\sqrt{3} - 1} = \frac{-\sqrt{3}\left(\frac{4}{3}\right)}{\sqrt{3} - 1}
\]

\[
= \frac{-\frac{4\sqrt{3}}{3}}{\sqrt{3} - 1}
\]

Thus, the exact value of \(\tan\left(\frac{5\pi}{6} + \frac{3\pi}{4}\right)\) is \(\frac{-4\sqrt{3}/3}{\sqrt{3} - 1}\).

Would you like further simplifications or do you have any questions?

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Here are 5 related questions:

1. How do you simplify expressions with square roots in the denominator?
2. What is the geometric interpretation of the tangent addition formula?
3. How does the sign of the tangent function vary across the four quadrants?
4. Can you explain why \(\tan(\pi - \theta) = -\tan(\theta)\)?
5. How would you find the exact value of \(\tan(A - B)\)?

**Tip:** When dealing with trigonometric functions of angles in different quadrants, always consider the sign of the function in that quadrant.

Discover the exact value of tan((5pi/6) + (3pi/4)) using the tangent addition formula. Follow our detailed step-by-step solution to understand trigonometric identities and calculations involving the tangent function.

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