Bài 111787

Question

Chứng minh rằng trong tam giác $ABC$ ta có hệ thức:
a) $\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1 (1)$
b) $\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2} (2)$

score 0 · Answer 1

a)
Ta có: $\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$
$\Rightarrow \tan \left (\frac{A+B}{2} \right )=\cot \frac{C}{2}=\frac{1}{\tan \frac{C}{2}}$
$\Leftrightarrow \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan\frac{A}{2}\tan\frac{B}{2}}=\frac{1}{\tan\frac{C}{2}}$
$\Rightarrow 1-\tan\frac{A}{2}\tan\frac{B}{2}=\tan\frac{C}{2}\left ( \tan \frac{A}{2}+\tan \frac{B}{2} \right )$
Ta suy ra đpcm.

b)
Ta có: $\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$
$\Rightarrow \tan \left (\frac{A+B}{2} \right )=\cot \frac{C}{2}=\frac{1}{\tan \frac{C}{2}}$
$\Leftrightarrow \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan\frac{A}{2}\tan\frac{B}{2}}=\frac{1}{\tan\frac{C}{2}}$
$\Leftrightarrow \frac{\frac{1}{\cot \frac{A}{2}}+\frac{1}{\cot\frac{B}{2}}}{1-\frac{1}{\cot\frac{A}{2}\cot\frac{B}{2}}}=\cot\frac{C}{2}$
$\Leftrightarrow \frac{\cot \frac{A}{2}+\cot \frac{B}{2}}{\cot\frac{A}{2}\cot\frac{B}{2}-1}=\cot\frac{C}{2}$
$\Rightarrow \cot \frac{A}{2}+\cot \frac{B}{2}=\cot\frac{A}{2}\cot\frac{B}{2}\tan\frac{C}{2}-\cot\frac{C}{2}$
Ta được đpcm.

Bài 111787

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Bài 111787

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