Example 14 Given $\cos \alpha=\tan \beta, \cos \beta=\tan \gamma, \cos \gamma=\tan \alpha$, then $\sin ^{2} \alpha=\sin ^{2} \beta=\sin ^{2} \gamma=\cos ^{4} \alpha=$ $\cos ^{4} \beta=\cos ^{4} \gamma=4 \sin ^{2} 18^{\circ}$.
Prove that if $x=\cos \alpha, y=\cos \beta, z=\cos \gamma$, then
$$
\begin{array}{c}
x^{2} y^{2}=\tan ^{2} \beta \cdot \cos ^{2} \beta=\sin ^{2} \beta=1-y^{2} \cdots \cdots \text { (1) } \\
y^{2} z^{2}=\tan ^{2} \gamma \cos ^{2} \gamma=\sin ^{2} \gamma=1-z^{2} \cdots \cdots \text { (2) } \\
z^{2} x^{2}=\tan ^{2} \alpha \cdot \cos ^{2} \alpha=\sin ^{2} \alpha=1-x^{2} \cdots \cdots \text { (3) }
\end{array}
$$

From (1): $y^{2}=\frac{1}{1+x^{2}} \cdots \cdots$ (4)
Substituting (4) into (2): $z^{2}=\frac{2+x^{2}}{1+x^{2}} \cdots \cdots$
Substituting (5) into (3): $2 x^{4}+2 x^{2}-2=0$,

Thus, $x^{2}=\frac{\sqrt{5}-1}{2}$, hence $x^{2}=y^{2}=z^{2}=\frac{\sqrt{5}-1}{2}=2 \cdot \sin 18^{\circ}$, so $\sin ^{2} \alpha=\sin ^{2} \beta \sin ^{2} \gamma=1-x^{2}=\left(\frac{\sqrt{5}-1}{2}\right)^{2}=4 \sin ^{2} 18^{\circ}$.