1. We are asked to prove the identity: $$\frac{\tan \alpha - \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\tan \alpha + \tan \beta}{1 + \tan \alpha \tan \beta}$$ 2. Recall the tangent subtraction and addition formulas: $$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$$ $$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$$ Note the denominators differ from the given expression. 3. Let's analyze the left side (LHS): $$LHS = \frac{\tan \alpha - \tan \beta}{1 - \tan \alpha \tan \beta}$$ 4. The right side (RHS) is: $$RHS = \frac{\tan \alpha + \tan \beta}{1 + \tan \alpha \tan \beta}$$ 5. Multiply numerator and denominator of LHS by $-1$ to rewrite denominator: $$LHS = \frac{\tan \alpha - \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{-(\tan \beta - \tan \alpha)}{-(1 - \tan \alpha \tan \beta)} = \frac{\tan \beta - \tan \alpha}{\tan \alpha \tan \beta - 1}$$ This does not simplify to RHS directly. 6. Alternatively, test with specific values to check equality: Let $\alpha = 45^\circ$, $\beta = 0^\circ$: $$\tan 45^\circ = 1, \tan 0^\circ = 0$$ LHS: $$\frac{1 - 0}{1 - 1 \cdot 0} = \frac{1}{1} = 1$$ RHS: $$\frac{1 + 0}{1 + 1 \cdot 0} = \frac{1}{1} = 1$$ They are equal for this case. 7. However, the original identity as stated is not a standard trigonometric identity and does not hold generally. 8. Conclusion: The given equality is false in general and cannot be proven as an identity. Final answer: The identity is not true for all $\alpha, \beta$ and thus cannot be proven.