1. **Problem statement:** Prove that $$\frac{(1 + \tan\theta)^n}{(1 - \tan\theta)^n} = \frac{1 + \tan\theta}{1 - \tan\theta}$$ only if $n=1$. 2. **Formula and rules:** The given expression is a ratio raised to the power $n$. For the equality to hold, the powers must be equal or the base must be 1. 3. **Step-by-step solution:** - Start with the equation: $$\frac{(1 + \tan\theta)^n}{(1 - \tan\theta)^n} = \frac{1 + \tan\theta}{1 - \tan\theta}$$ - Rewrite the left side as: $$\left(\frac{1 + \tan\theta}{1 - \tan\theta}\right)^n = \frac{1 + \tan\theta}{1 - \tan\theta}$$ - Since the bases are the same, for the equality to hold, the exponents must be equal: $$n = 1$$ 4. **Explanation:** - The expression is a power equality with the same base. - The only way for $$a^n = a$$ (where $$a \neq 0,1$$) is if $$n=1$$. 5. **Final answer:** $$\boxed{n=1}$$