Question 44, Answer: Let $\mathrm{f}(\mathrm{x})=\frac{1}{2} \sin 2 \mathrm{x} \tan \mathrm{x}+2 \sin \mathrm{x} \tan \frac{\mathrm{x}}{2}$, then $\cos \mathrm{x} \neq 0, \cos \frac{\mathrm{x}}{2} \neq 0$, which means $\cos x \neq 0, -1$. Because
$$
\begin{array}{l}
\mathrm{f}(\mathrm{x})=\frac{1}{2} \sin 2 \mathrm{x} \tan \mathrm{x}+2 \sin x \tan \frac{x}{2}=\sin ^{2} \mathrm{x}+2 \sin \mathrm{x} \cdot \frac{1-\cos \mathrm{x}}{\sin \mathrm{x}}=\sin ^{2} \mathrm{x}+2(1-\cos \mathrm{x}) \\
=4-(1+\cos \mathrm{x})^{2}
\end{array}
$$

Therefore, the range of $f(x)$ is $[0,3) \cup(3,4)$.
Question 44, Function $\mathrm{y}=\frac{1}{2} \sin 2 \mathrm{x} \tan \mathrm{x}+2 \sin \mathrm{x} \tan \frac{\mathrm{x}}{2}$ has the range $\qquad$