Question

[Mi Phan]

cota/2+cotb/2+cotc/2=cota/2.cotb/2.cotc/2

Answer 1

[Nguyễn Việt Lâm]

Chỉ đúng với điều kiện A, B, C là 3 góc trong tam giác \(\Rightarrow A+B+C=\pi\)

Đặt \(\frac{A}{2}=x\) , \(\frac{B}{2}=y\); \(\frac{C}{2}=z\) \(\Rightarrow x+y+z=\frac{\pi}{2}\Rightarrow\left\{{}\begin{matrix}x+y=\frac{\pi}{2}-z\\z=\frac{\pi}{2}-\left(x+y\right)\end{matrix}\right.\)

\(cot\frac{A}{2}+cot\frac{B}{2}+cot\frac{C}{2}=cotx+coty+cotz=\frac{cosx}{sinx}+\frac{cosy}{siny}+\frac{cosz}{sinz}\)

\(=\frac{cosx.siny+cosy.sinx}{sinx.siny}+\frac{cosz}{sinz}=\frac{sin\left(x+y\right)}{sinx.siny}+\frac{cosz}{sinz}\)

\(=\frac{sin\left(\frac{\pi}{2}-z\right)}{sinx.siny}+\frac{cosz}{sinz}=\frac{cosz}{sinx.siny}+\frac{cosz}{sinz}=cosz\left(\frac{1}{sinx.siny}+\frac{1}{sinz}\right)\)

\(=\frac{cosz}{sinx.siny.sinz}\left(sinz+sinx.siny\right)=\frac{cosz}{sinx.siny.sinz}\left(sin\left(\frac{\pi}{2}-\left(x+y\right)\right)+sinxsiny\right)\)

\(=\frac{cosz}{sinx.siny.sinz}\left(cos\left(x+y\right)+sinx.siny\right)\)

\(=\frac{cosz}{sinx.siny.sinz}\left(cosx.cosy-sinx.siny+sinx.siny\right)\)

\(=\frac{cosx.cosy.cosz}{sinx.siny.sinz}=cotx.coty.cotz=cot\frac{A}{2}.cot\frac{B}{2}.cot\frac{C}{2}\)