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The tangent function, often denoted as \( \tan(\theta) \), plays a crucial role in trigonometry. It is defined as the ratio of the sine and cosine functions: \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). This definition means that for any angle \( \theta \), the tangent function gives us the ratio of the lengths of the opposite side to the adjacent side in a right-angled triangle.
Understanding this relationship helps solve numerous trigonometric problems, especially those involving right triangles or angles in the unit circle.
Using the tangent function allows mathematicians to find missing side lengths and angles, making it a cornerstone in geometry and trigonometry courses.

• Tangent is undefined when \( \cos(\theta) = 0 \), because division by zero is impossible. These angles are typically odd multiples of \( \frac{\pi}{2} \) (90 degrees).
• The tangent function has a periodicity of \( \pi \), meaning \( \tan(\theta + \pi) = \tan(\theta) \).

In trigonometry, 'special angles' refer to certain commonly used angles for which trigonometric functions have known, simple values. These angles include \( 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} \), and their multiples.
Understanding these angles aids in solving trigonometric problems without using a calculator, because their sine, cosine, and tangent values are easily memorized and frequently appear in problems.
The angle \( \frac{\pi}{6} \), which equates to 30 degrees, is essential in problems like the one we are exploring. For this angle:

• \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \)
• \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)
• \( \tan\left(\frac{\pi}{6}\right) = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\sqrt{3}} \)

Special angles and their values help in sketching and understanding trigonometric graphs too.

Rationalizing the denominator is a mathematical process used to eliminate radicals from the denominator of a fraction.
In our exercise, after finding \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \), we rationalize the denominator for a cleaner, conventional form of the result. This is often preferred in mathematics for ease of calculation and comparison.
To rationalize \( \frac{1}{\sqrt{3}} \), multiply both numerator and denominator by \( \sqrt{3} \):

• You obtain \( \frac{1 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{\sqrt{3}}{3} \).
• This process results in a rationalized form, free of square roots in the denominator.

Rationalizing is essential in mathematics, particularly in simplifying expressions and making them easier to understand and work with.