Question

Chứng minh:

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

cotA.cotB + cotB.cotC + cotC.cotA = 1

Answer 1

\(A+B+C=180^0\Rightarrow\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^0\Rightarrow\frac{A}{2}+\frac{B}{2}=90^0-\frac{C}{2}\)

\(\Rightarrow tan\left(\frac{A}{2}+\frac{B}{2}\right)=tan\left(90^0-\frac{C}{2}\right)\)

\(\Leftrightarrow\frac{tan\frac{A}{2}+tan\frac{B}{2}}{1-tan\frac{A}{2}.tan\frac{B}{2}}=cot\frac{C}{2}=\frac{1}{tan\frac{C}{2}}\)

\(\Leftrightarrow tan\frac{C}{2}\left(tan\frac{A}{2}+tan\frac{B}{2}\right)=1-tan\frac{A}{2}.tan\frac{B}{2}\)

\(\Leftrightarrow tan\frac{A}{2}tan\frac{C}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{A}{2}.tan\frac{B}{2}=1\)

b/\(A+B+C=180^0\Rightarrow A+B=180^0-C\)

\(\Rightarrow cot\left(A+B\right)=cot\left(180^0-C\right)\)

\(\Leftrightarrow\frac{cotA.cotB-1}{cotA+cotB}=-cotC\)

\(\Leftrightarrow cotA.cotB-1=-cotA.cotC-cotB.cotC\)

\(\Leftrightarrow cotA.cotB+cotB.cotC+cotA.cotC=1\)