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To solve \( \tan(-2700^\circ) \), we first need to simplify the angle to a standard position, as the tangent function is periodic.

### Step 1: Reduce the angle
The period of the tangent function is \( 180^\circ \), so we can reduce the angle \( -2700^\circ \) by finding its equivalent angle within \( [0^\circ, 360^\circ] \).

We perform:
\[
-2700^\circ \div 360^\circ = -7.5
\]
This means that \( -2700^\circ \) is equivalent to 7 and a half full rotations counterclockwise.

Next, we find the remainder when dividing by \( 360^\circ \):
\[
-2700^\circ + (8 \times 360^\circ) = -2700^\circ + 2880^\circ = 180^\circ
\]
Thus, \( \tan(-2700^\circ) \) is equivalent to \( \tan(180^\circ) \).

### Step 2: Evaluate the tangent at \( 180^\circ \)
We know that:
\[
\tan(180^\circ) = 0
\]

Thus, the exact value of \( \tan(-2700^\circ) \) is \( 0 \).

Would you like further details, or do you have any additional questions?

Here are 5 related questions to expand on this:
1. What is the period of the sine and cosine functions?
2. How do you reduce large angles in other trigonometric functions?
3. Why does the tangent function repeat every \( 180^\circ \)?
4. What is the tangent of other common angles like \( 90^\circ \) or \( 270^\circ \)?
5. How does the tangent function behave with negative angles?

**Tip**: When dealing with large or negative angles, reducing them to a standard range (usually \( 0^\circ \) to \( 360^\circ \)) makes calculations easier.

Discover how to find the exact value of tan(-2700°) by reducing the angle using the periodic nature of the tangent function. This problem involves key trigonometric concepts such as angle reduction and periodic functions, suitable for students in Grades 10-12.

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