Example 3 Simplify $\frac{1+\sin \theta+\cos \theta}{1+\sin \theta-\cos \theta}+\frac{1-\cos \theta+\sin \theta}{1+\cos \theta+\sin \theta}$.
Solution 1: Since $\frac{1+\sin \theta+\cos \theta}{1+\sin \theta-\cos \theta}=\frac{2 \cos ^{2} \frac{\theta}{2}+2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}+2 \sin \frac{\theta}{2}+\cos \frac{\theta}{2}}$
$$
=\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}=\operatorname{ctg} \frac{\theta}{2}
$$

Therefore, the original expression $=\operatorname{ctg} \frac{\theta}{2}+\operatorname{tg} \frac{\theta}{2}=\frac{\sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}=2 \csc \theta$.
Solution 2: The original expression $=\frac{(1+\sin \theta-\cos \theta)^{2}+(1+\sin \theta+\cos \theta)^{2}}{(1+\sin \theta)^{2}-\cos ^{2} \theta}$
$$
\begin{array}{l}
=\frac{2(1+\sin \theta)^{2}+2 \cos ^{2} \theta}{2 \sin ^{2}+2 \sin ^{2} \theta} \\
=\frac{2}{\sin \theta}=2 \csc \theta
\end{array}
$$

Solution 3: Since $\operatorname{tg} \frac{\theta}{2}=\frac{1-\cos \theta}{\sin \theta}=\frac{\sin \theta}{1+\cos \theta}$, by the principle of proportion, we get $\operatorname{tg} \frac{\theta}{2}=\frac{1-\cos \theta+\sin \theta}{\sin \theta+1+\cos \theta}$. Therefore, the original expression $=\operatorname{ctg} \frac{\theta}{2}+\operatorname{tg} \frac{\theta}{2}=2 \csc \theta$.
Solution 4: Since $\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}=\frac{\frac{1-\cos \theta}{\sin \theta}-1}{\frac{1+\cos \theta}{\sin \theta}+1}$
$$
=\frac{\operatorname{tg} \frac{\theta}{2}+1}{\operatorname{ctg} \frac{\theta}{2}+1}=\operatorname{tg} \frac{\theta}{2} \text {. }
$$

Therefore, the original expression $=\operatorname{ctg} \frac{\theta}{2}+\operatorname{tg} \frac{\theta}{2}=2 \csc \theta$.