Prove: $\frac{\sin ^{2 n} \theta}{a^{n-1}}+\frac{\cos ^{2 n} \theta}{b^{n-1}}=\frac{1}{(a+b)^{n-1}}$.
Proof: By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l} 1=(a+b)\left(\frac{\sin ^{4} \theta}{a}+\frac{\cos ^{4} \theta}{b}\right) \\ \geqslant\left(\sqrt{a} \cdot \frac{\sin ^{2} \theta}{\sqrt{a}}+\sqrt{b} \cdot \frac{\cos ^{2} \theta}{\sqrt{b}}\right)^{2} \\ =\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}=1 \end{array}$$

Furthermore, by the condition for equality in the Cauchy-Schwarz inequality, we get
$$\frac{\sin ^{2} \theta}{a}=\frac{\cos ^{2} \theta}{b}=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{a+b}=\frac{1}{a+b} .$$

Thus, $\sin ^{2} \theta=\frac{a}{a+b}, \cos ^{2} \theta=\frac{b}{a+b}$.
Therefore, $\frac{\sin ^{2 n} \theta}{a^{n-1}}+\frac{\cos ^{2 n} \theta}{b^{n-1}}$
$$\begin{array}{l} =\left(\frac{a}{a+b}\right)^{n} \frac{1}{a^{n-1}}+\left(\frac{b}{a+b}\right)^{n} \frac{1}{b^{n-1}} \\ =\frac{a+b}{(a+b)^{n}}=\frac{1}{(a+b)^{n-1}} \end{array}$$

Note: The special aspect of this problem is that, for given positive numbers $a$ and $b$, the expression $\frac{\sin ^{4} \theta}{a}+\frac{\cos ^{4} \theta}{b}$ achieves its maximum value $\frac{1}{a+b}$, which uniquely determines $\sin ^{2} \theta$ and $\cos ^{2} \theta$:
$$\sin ^{2} \theta=\frac{a}{a+b}, \cos ^{2} \theta=\frac{b}{a+b}$$