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回復 1# qaz 的帖子

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回復 3# bch0722b 的帖子

\(\begin{align}
  & A+B+C=\pi  \\
& \tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}+\tan \frac{A}{2}\tan \frac{B}{2} \\
& \text{=}\tan \frac{C}{2}\left( \tan \frac{B}{2}+\tan \frac{A}{2} \right)\text{+}\tan \frac{A}{2}\tan \frac{B}{2} \\
& \text{=}\tan \frac{C}{2}\tan \left( \frac{B}{2}\text{+}\frac{A}{2} \right)\left( 1-\tan \frac{B}{2}\tan \frac{A}{2} \right)\text{+}\tan \frac{A}{2}\tan \frac{B}{2} \\
& \text{=}\tan \frac{C}{2}\tan \left( \frac{\pi }{2}-\frac{C}{2} \right)\left( 1-\tan \frac{B}{2}\tan \frac{A}{2} \right)\text{+}\tan \frac{A}{2}\tan \frac{B}{2} \\
& =1-\tan \frac{B}{2}\tan \frac{A}{2}+\tan \frac{A}{2}\tan \frac{B}{2} \\
& =1 \\
\end{align}\)


\(\begin{align}
  & x=\tan \frac{B}{2}\tan \frac{C}{2},y=\tan \frac{C}{2}\tan \frac{A}{2},z=\tan \frac{A}{2}\tan \frac{B}{2} \\
& 1+\cos A=1+\frac{1-{{\tan }^{2}}\frac{A}{2}}{1+{{\tan }^{2}}\frac{A}{2}}=\frac{2}{1+{{\tan }^{2}}\frac{A}{2}} \\
& \frac{1+\cos A}{2}=\frac{1}{1+{{\tan }^{2}}\frac{A}{2}}=\frac{1}{1+\frac{yz}{x}}=\frac{x}{x+yz} \\
&  \\
& \cos \frac{A}{2}=\sqrt{\frac{1+\cos A}{2}}=\sqrt{\frac{x}{x+yz}} \\
\end{align}\)

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