Cho tam giác ABC, khi đó AB=c; Ac=b; BC=a;[tex] \widehat{BAC}=A [/tex]; [tex] \widehat{ABC}=B [/tex]; [tex]\widehat{ACB}=C[/tex]
CM đẳng thức
f)tanA+tanB+tanC=tanA.tanB.tanC
g)[tex]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[/tex]
h)cotgA cotgB+cotgB cotg C+cotgC cotgA=1
i)[tex]cotg.\frac{A}{2}+cotg.\frac{B}{2}+cotg.\frac{C}{2}=cotg.\frac{A}{2}cotg.\frac{B}{2}cotg.\frac{C}{2}[/tex]
[TEX]f) tan(A+B)=\frac{tanA+tanB}{1-tanA.tanB}=tan(\pi -C)=-tanC \Rightarrow tanA+tanB=tanA.tanB.tanC-tanC[/TEX]
g)[TEX]\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2} \Rightarrow tan \frac{A}{2}.tan (\frac{B}{2}+\frac{C}{2})=1 [/TEX]
[TEX]\Rightarrow tan \frac{A}{2}=\frac{1}{tan (\frac{B}{2}+\frac{C}{2})}=\frac{1-tan \frac{B}{2}.tan \frac{C}{2}}{tan \frac{B}{2}+tan \frac{C}{2}} \Rightarrow tan \frac{A}{2}(tan \frac{B}{2} + tan \frac{C}{2})=1-tan \frac{B}{2}.tan \frac{C}{2} [/TEX]
h) [TEX]tanA+tanB+tanC=tanA.tanB.tanC \Leftrightarrow \frac{tanA+tanB+tanC}{tanA.tanB.tanC}=1 \Leftrightarrow \frac{1}{tanA.tanB}+\frac{1}{tanA.tanC}+\frac{1}{tanB.tanC}=1[/TEX]
[TEX]\Rightarrow cotgA cotgB+cotgB cotg C+cotgC cotgA=1[/TEX]
i) [tex]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[/tex]
[tex]\Leftrightarrow \frac{tan \frac{A}{2}.tan \frac{B}{2}}{tan \frac{A}{2}.tan \frac{B}{2}.tan \frac{C}{2}}+\frac{tan \frac{A}{2}.tan \frac{C}{2}}{tan \frac{A}{2}.tan \frac{B}{2}.tan \frac{C}{2}}+\frac{tan \frac{B}{2}.tan \frac{C}{2}}{tan \frac{A}{2}.tan \frac{B}{2}.tan \frac{C}{2}}=\frac{1}{tan \frac{A}{2}tan \frac{B}{2}tan \frac{C}{2}[/tex]
[TEX]\Rightarrow \frac{1}{tan \frac{A}{2}}+\frac{1}{tan \frac{B}{2}}+\frac{1}{tan \frac{C}{2}}=\frac{1}{tan \frac{A}{2}tan \frac{B}{2}tan \frac{C}{2}} \Leftrightarrow cotg.\frac{A}{2}+cotg.\frac{B}{2}+cotg.\frac{C}{2}=cotg\frac{A}{2}.cotg\frac{B}{2}.cotg\frac{C}{2}
[/TEX]