Example 4 Real numbers $x, y$ satisfy
$$
\frac{(x+2)^{2}}{9}+\frac{(y-2)^{2}}{16}=1 \text {. }
$$

If $x+y-k<0$ always holds, find the range of $k$. Analysis: From $\frac{(x+2)^{2}}{9}+\frac{(y-2)^{2}}{16}=1$, we know $-1 \leqslant \frac{x+2}{3} \leqslant 1,-1 \leqslant \frac{y-2}{4} \leqslant 1$.
Therefore, trigonometric substitution can be used.
Solution: Let $\frac{x+2}{3}=\cos \theta, \frac{y-2}{4}=\sin \theta$, where $\theta \in \mathbf{R}$, then
$$
\begin{array}{l}
x=3 \cos \theta-2, y=4 \sin \theta+2 . \\
\text { Substituting } x+y-k3 \cos \theta+4 \sin \theta=5 \sin (\theta+\varphi)$, where, $\tan \varphi=\frac{3}{4}$.
Since $\theta \in \mathbf{R}$, then $5 \sin (\theta+\varphi) \in[-5,5]$.
Therefore, for $k>5 \sin (\theta+\varphi)$ to always hold, it is necessary and sufficient that $k>5$, i.e., $k \in(5,+\infty)$.

Comment: Examples 2, 3, and 4 all belong to the category where the given conditions conform to the basic trigonometric identity $\cos ^{2} \theta+\sin ^{2} \theta=1$. Additionally, there are other basic identities: $\sec ^{2} \theta-\tan ^{2} \theta=1, \csc ^{2} \theta$ $-\cot ^{2} \theta=1$, etc.