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

The equation you've provided is the formula for the tangent of the sum of two angles, \( A \) and \( B \). This formula is derived from the definitions of the tangent function in terms of sine and cosine. To understand this formula, let's break it down step by step: 1. **Understanding Tangent**: The tangent of an angle is defined as the ratio of the sine of that angle to the cosine of that angle: \[ \tan A = \frac{\sin A}{\cos A}, \quad \tan B = \frac{\sin B}{\cos B} \] 2. **Using the Angle Addition Formula**: The sine and cosine of the sum of two angles can be expressed as: \[ \sin(A + B) = \sin A \cos B + \cos A \sin B \] \[ \cos(A + B) = \cos A \cos B - \sin A \sin B \] 3. **Finding \( \tan(A + B) \)**: Using the definitions of sine and cosine, we can express \( \tan(A + B) \) as: \[ \tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)} = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B} \] 4. **Substituting \( \tan A \) and \( \tan B \)**: Now, substituting \( \tan A \) and \( \tan B \) into the equation: - Let \( x = \tan A \) and \( y = \tan B \). - Then, \( \sin A = \frac{x}{\sqrt{1+x^2}} \) and \( \cos A = \frac{1}{\sqrt{1+x^2}} \). - Similarly, for \( B \). 5. **Simplifying the Expression**: After substituting and simplifying, we arrive at: \[ \tan(A + B) = \frac{x + y}{1 - xy} \] Thus, the formula \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \) is confirmed. If you have any specific questions or need further clarification on any part of this derivation, feel free to ask!