Trong mọi tam giác ABC, ta có: \(\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}=\frac{\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2}}{k \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}\). Khi đó k bằng với
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Lời giải:
Báo saiTa có:
\(\begin{aligned} =&\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\\ =& \frac{1}{2}\left[\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)+\left(\tan \frac{B}{2}+\tan \frac{C}{2}\right)+\left(\tan \frac{C}{2}+\tan \frac{A}{2}\right)\right] \\ =& \frac{1}{2}\left[\frac{\sin \frac{A+B}{2}}{\cos \frac{A}{2} \cos \frac{B}{2}}+\frac{\sin \frac{B+C}{2}}{\cos \frac{B}{2} \cos \frac{C}{2}}+\frac{\sin \frac{C+A}{2}}{\cos \frac{C}{2} \cos \frac{A}{2}}\right] \\ =& \frac{\cos ^{2} \frac{C}{2}+\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}}{2 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} \end{aligned}\)
Vậy k=2.