Bài 106895

Question

Chứng minh rằng trong mọi tam giác ABC ta luôn có:
$\tan \frac{A}{4} + \tan \frac{B}{4} + \tan \frac{C}{4} + \tan \frac{A}{4}\tan \frac{B}{4} + \tan \frac{B}{4}\tan \frac{C}{4} $
$+ \tan \frac{A}{4}\tan \frac{C}{4} - \tan \frac{A}{4}\tan \frac{B}{4}\tan \frac{C}{4} = 1$

score 0 · Answer 1

Ta có:
$
\tan \frac{{A + B}}{4} = \tan \frac{{\pi - C}}{4}\,\,hay\,\frac{{\tan \frac{A}{4} + \tan
\frac{B}{4}}}{{1 - \tan \frac{A}{4}\tan \frac{B}{4}}} = \frac{{1 - \tan \frac{C}{4}}}{{1 + \tan
\frac{C}{4}}}\\
\Leftrightarrow \tan \frac{A}{4} + \tan \frac{B}{4} + \tan \frac{A}{4}\tan \frac{C}{4} + \tan
\frac{B}{4}\tan \frac{C}{4} $
$= 1 - \tan \frac{C}{4} - \tan \frac{A}{4}\tan \frac{B}{4} + \tan
\frac{C}{4}\\
\Rightarrow \tan \frac{A}{4} + \tan \frac{B}{4} + \tan \frac{C}{4} + \tan \frac{A}{4}\tan
\frac{B}{4} $
$+ \tan \frac{B}{4}\tan \frac{C}{4} + \tan \frac{A}{4}\tan \frac{B}{4} - \tan
\frac{A}{4}\tan \frac{B}{4}\tan \frac{C}{4} = 1
$

Bài 106895

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Bài 106895

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