We are tasked with finding the value of:

\[
\cot 15^\circ - \tan 15^\circ
\]

We can use the identities for cotangent and tangent to solve this:

\[
\cot 15^\circ = \frac{1}{\tan 15^\circ}
\]
\[
\tan 15^\circ = \tan(45^\circ - 30^\circ)
\]

The tangent of a difference formula is:

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

Applying this to \(\tan 15^\circ\):

\[
\tan 15^\circ = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \cdot \tan 30^\circ}
\]

We know the values:

\[
\tan 45^\circ = 1, \quad \tan 30^\circ = \frac{1}{\sqrt{3}}
\]

Substitute these into the equation:

\[
\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}}
\]

Simplifying:

\[
\tan 15^\circ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}
\]

Thus:

\[
\cot 15^\circ = \frac{1}{\tan 15^\circ} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}
\]

Now, we compute \(\cot 15^\circ - \tan 15^\circ\):

\[
\cot 15^\circ - \tan 15^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} - \frac{\sqrt{3} - 1}{\sqrt{3} + 1}
\]

We can simplify this by finding a common denominator and subtracting the two fractions.

Using the difference of squares:

\[
(\sqrt{3} - 1)(\sqrt{3} + 1) = \sqrt{3}^2 - 1^2 = 3 - 1 = 2
\]

Thus:

\[
\cot 15^\circ - \tan 15^\circ = \frac{(\sqrt{3} + 1)^2 - (\sqrt{3} - 1)^2}{2}
\]

Now compute the squares:

\[
(\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}
\]
\[
(\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}
\]

So:

\[
\cot 15^\circ - \tan 15^\circ = \frac{(4 + 2\sqrt{3}) - (4 - 2\sqrt{3})}{2} = \frac{4\sqrt{3}}{2} = 2\sqrt{3}
\]

Thus, the value is:

\[
\boxed{2\sqrt{3}}
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. What is the general formula for \(\cot(A - B)\)?
2. How do you find \(\tan 75^\circ\) using the same identity for tangent?
3. Can you derive \(\tan 30^\circ\) geometrically?
4. How does the tangent of small angles like 15° behave compared to large angles?
5. Why is the cotangent function the reciprocal of tangent?

**Tip:** Memorizing common trigonometric values like \(\tan 30^\circ\) and \(\tan 45^\circ\) simplifies these calculations!

Explore how to find the value of cot 15° - tan 15° using trigonometric identities and the tangent difference formula. This problem is ideal for students in Grades 10-12, covering key trigonometric concepts, such as cotangent, tangent, and the difference of squares.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation