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已知锐角三角形 ${\displaystyle ABC}$ 的内角 ${\displaystyle A,B,C}$ 的对边分别为 ${\displaystyle a,b,c}$, 且 ${\displaystyle a^2+b^2+c^2=4ab}$.

1. 求角 ${\displaystyle C}$ 的取值范围;
2. 证明: ${\displaystyle \cos (A-B)=\frac{{\cos }^{2}C+2\cos C-2}{\cos C-2}}$;
3. 求 ${\displaystyle \frac{1}{{\tan }^{2}A}+\frac{1}{{\tan }^{2}B}+\frac{3}{{\tan }^{2}C}}$ 的取值范围.

1. 解 首先 ${\displaystyle C\in (0,\frac{\pi }{2})}$. 不妨设 ${\displaystyle a\ge b}$. 则在锐角三角形中 ${\displaystyle b^2+c^2>a^2}$, 也即 ${\displaystyle 4ab-a^2>a^2}$, 解得 ${\displaystyle a<2b}$. 所以 ${\displaystyle 1\le \frac{b}{a}<2}$. 这样根据余弦定理 $$\begin{array}{}\text{(1)}& \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{a}{b}+\frac{b}{a}-2\in [0,\frac{1}{2}).\end{array}$$ 这样 ${\displaystyle C\in (\frac{\pi }{3},\frac{\pi }{2})}$.
2. 证明 根据 (1) 和正弦定理 $$\begin{array}{rl}\text{(2)}& \cos C+2& =\frac{{\sin }^{2}A+{\sin }^{2}B}{\sin A\sin B}& =\frac{2-\cos 2A-\cos 2B}{\cos (A-B)-\cos (A+B)}\\ & =\frac{2-2\cos (A+B)\cos (A-B)}{\cos (A-B)-\cos (A+B)}\\ & =\frac{2+2\cos C\cos (A-B)}{\cos (A-B)+\cos C},\end{array}$$ 解得 $$\begin{array}{}\text{(3)}& \cos (A-B)=\frac{{\cos }^{2}C+2\cos C-2}{\cos C-2}.\end{array}$$
3. 解 $$\begin{array}{rl}& \frac{1}{{\tan }^{2}A}+\frac{1}{{\tan }^{2}B}=\frac{1}{{\sin }^{2}A}+\frac{1}{{\sin }^{2}B}-2\\ =& \frac{{\sin }^{2}A+{\sin }^{2}B}{{\sin }^{2}A{\sin }^{2}B}-2\stackrel{(1)}{=}\frac{\cos C+2}{\sin A\sin B}-2\\ =& \frac{2(\cos C+2)}{\cos (A-B)-\cos (A+B)}-2\\ \stackrel{(2)}{=}& \frac{2(\cos C+2)}{{\displaystyle \frac{{\cos }^{2}C+2\cos C-2}{\cos C-2}}+\cos C}-2\\ =& \frac{{\cos }^{2}C-4}{{\cos }^{2}C-1}-2=-1+\frac{3}{{\sin }^{2}C},\end{array}$$ 于是$$\begin{array}{r}\frac{1}{{\tan }^{2}A}+\frac{1}{{\tan }^{2}B}+\frac{3}{{\tan }^{2}C}=\frac{6}{{\sin }^{2}C}-4\in (2,4).\end{array}$$