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题：在三角形 $\triangle ABC$ 中满足： $\frac {a}{\cos^2\frac A2}=\frac {b}{2\cos^2\frac B2}=\frac {c}{3\cos^2\frac C2},$ 则 $\tan B=$ _______.
由正弦定理，
$\begin{gathered} \frac {a}{\cos^2\frac A2}=\frac {b}{2\cos^2\frac B2}=\frac {c}{3\cos^2\frac C2}\\[1em] \frac {\sin A}{\cos^2\frac A2}=\frac {\sin B}{2\cos^2\frac B2}=\frac {\sin C}{3\cos^2\frac C2}\\[1em] \frac {2\sin \frac A2\cos\frac A2}{\cos^2\frac A2}=\frac {2\sin \frac B2\cos\frac B2}{2\cos^2\frac B2}=\frac {2\sin \frac C2\cos\frac C2}{3\cos^2\frac C2}\\[1em] \tan \frac A2=\frac 12 \tan \frac B2=\frac 13 \tan \frac C2\\[1em] \therefore \ \ \tan \frac B2=2\tan \frac A2\\[1em] \ \ \tan \frac C2=3\tan \frac A2 \end{gathered}$
另一方面在 $\triangle ABC$ 中： $A=\pi -(B+C)$ ，则
$\begin{gathered} \color{red}{\frac {1}{\tan \frac A2}}=\frac {1}{\tan \frac{\pi-(B+C)}2}=\frac {1}{\tan \left(\frac \pi2-\frac{B+C}2\right)}\\[1em] =\tan\frac{B+C}2=\tan \left(\frac B2+\frac C2\right) =\color{red}{\frac{\tan\frac B2+\tan\frac C2}{1-\tan\frac B2\tan\frac C2}}\\[1.5em] \therefore \ \ \tan \frac A2\tan \frac B2+ \tan \frac B2\tan \frac C2+ \tan \frac C2\tan \frac A2=1 \end{gathered}$
进步有
$\begin{gathered} \tan \frac A2\cdot 2\tan \frac A2+ 2\tan \frac B2\cdot 3\tan \frac A2+ 3\tan \frac A2\cdot \tan \frac A2=1\\[1em] \Rightarrow \quad \tan \frac A2=\frac 1{\sqrt {11}},\\[1em] \therefore \quad \tan \frac B2=\frac {2}{\sqrt {11}},\\[1em] \Rightarrow \quad \tan B=\frac {2\tan \frac B2}{1-\tan^2\frac B2}=\frac {4\sqrt {11}}7. \end{gathered}$ |
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