Proof: The proofs of these inequalities are largely similar, and here we only prove (10). We have
$$\begin{array}{l} \operatorname{tg}^{2} \frac{A}{2}+\operatorname{tg}^{2} \frac{B}{2}=\frac{4\left(\sin \frac{A+B}{2}\right)^{2}-2 \sin A \sin B}{\left(\cos \frac{A-B}{2}+\cos \frac{A+B}{2}\right)^{2}} \\ =\frac{-2\left(\cos \frac{A-B}{2}\right)^{2}+4 \cos ^{2} \frac{C}{2}+2 \sin ^{2} \frac{C}{2}}{\left(\cos \frac{A-B}{2}+\sin \frac{C}{2}\right)^{2}} \end{array}$$

Let $a=\sin \frac{C}{2}>0, b=4 \cos ^{2} \frac{C}{2}+2 \sin ^{2} \frac{C}{2}>0$, then the above expression becomes: $\operatorname{tg}^{2} \frac{A}{2}+\operatorname{tg}^{2} \frac{B}{2}=\frac{-2\left(\cos \frac{A-B}{2}\right)^{2}+b}{\left(\cos \frac{A-B}{2}+a\right)^{2}}$
We can prove that the function $g(x)=\frac{-2 x^{2}+b}{(x+a)^{2}}(a>0, b>0)$ is a monotonically decreasing function on the interval $(0,+\infty)$.

Let $x_{1}>x_{2}>0$, then
$$\begin{array}{l} g\left(x_{2}\right)-g\left(x_{1}\right) \\ =\frac{2 a^{2}\left(x_{1}^{2}-x_{2}{ }^{2}\right)+4 a x_{1} x_{2}\left(x_{1}-x_{2}\right)}{\left(x_{1}+a\right)^{2}\left(x_{2}+a\right)^{2}} \\ \quad+b\left(x_{1}^{2}-x_{2}^{2}\right)+2 a b\left(x_{1}-x_{2}\right) \end{array}$$
$>0$, which means $g\left(x_{2}\right)>g\left(x_{1}\right)$.
Therefore, if in $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ we have $C=C^{\prime}$, then we can deduce:
$$\begin{array}{l} \operatorname{tg}^{2} \frac{A}{2}+\operatorname{tg}^{2} \frac{B}{2}>\operatorname{tg}^{2} \frac{A^{\prime}}{2}+\operatorname{tg}^{2} \frac{B^{\prime}}{2} \\ \Leftrightarrow \Rightarrow \cos \frac{A-B}{2}<\cos \frac{A^{\prime}-B^{\prime}}{2} \\ \Leftrightarrow\left|A^{\prime}-B^{\prime}\right|<|A-B| \end{array}$$

Thus, according to the corollary, we obtain inequality (10).