Prove the difference formula for tangent: Show that \( \tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)} \).

To prove the difference formula for tangent, which states that \[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}, \] we can utilize the fundamental definitions and identities of trigonometric functions. Here's a step-by-step proof: ### 1. **Start with the Definition of Tangent** The tangent of an angle is defined as the ratio of the sine to the cosine of that angle: \[ \tan(\theta) = \frac{\sin \theta}{\cos \theta} \] Therefore, the tangent of the difference of two angles \(a\) and \(b\) is: \[ \tan(a - b) = \frac{\sin(a - b)}{\cos(a - b)} \] ### 2. **Apply the Sine and Cosine Difference Identities** Recall the sine and cosine difference formulas: \[ \sin(a - b) = \sin a \cos b - \cos a \sin b \] \[ \cos(a - b) = \cos a \cos b + \sin a \sin b \] Substitute these into the expression for \(\tan(a - b)\): \[ \tan(a - b) = \frac{\sin a \cos b - \cos a \sin b}{\cos a \cos b + \sin a \sin b} \] ### 3. **Divide Numerator and Denominator by \(\cos a \cos b\)** To simplify the expression, divide both the numerator and the denominator by \(\cos a \cos b\): \[ \tan(a - b) = \frac{ \frac{\sin a \cos b}{\cos a \cos b} - \frac{\cos a \sin b}{\cos a \cos b} }{ \frac{\cos a \cos b}{\cos a \cos b} + \frac{\sin a \sin b}{\cos a \cos b} } \] Simplify each term: \[ \tan(a - b) = \frac{ \frac{\sin a}{\cos a} - \frac{\sin b}{\cos b} }{ 1 + \frac{\sin a \sin b}{\cos a \cos b} } \] ### 4. **Recognize the Tangent Definitions** Notice that \(\frac{\sin a}{\cos a} = \tan a\) and \(\frac{\sin b}{\cos b} = \tan b\). Also, \(\frac{\sin a \sin b}{\cos a \cos b} = \tan a \tan b\). Substitute these into the equation: \[ \tan(a - b) = \frac{ \tan a - \tan b }{ 1 + \tan a \tan b } \] ### 5. **Conclusion** Thus, we've derived the difference formula for tangent: \[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \] **Q.E.D.**