TA CÓ $cosA +cosB=1$$\Rightarrow 2.cos\frac{A+B}{2}.cos\frac{A-B}2=1$
$\Rightarrow sin\frac{C}2.cos\frac{A-B}2=\frac{1}2$
HAY $cos\frac{A-B}2=\frac1{2sin\frac{C}2}(1)$
MẶT KHÁC $tan\frac{A}2+tan\frac{B}2=\frac{sinA/2}{cosA/2}+\frac{sinB/2}{cosB/2}$
$ = \frac{sin(A+B)/2} {1/2[cos(A+B)/2+cos(A-B)/2]} $
$ = \frac{cosC/2}{1/2[sinC/2+cos(A-B)/2]} $
$ = \frac{2sinC}{2sin^2C/2+1} $
$\Rightarrow \frac{sinC}{2sin^2C/2+1}=\frac{1}{\sqrt3}$
$\Leftrightarrow \sqrt3sinC=2sin^2C/2+1=2-cosC$
$\Leftrightarrow 3(1-cos^2C)=(2-cosC)^2 \Leftrightarrow cosC=1/2$
$\Rightarrow \widehat{C}=60^0$
Thế điều trên vào (1),ta có $cos(A-B)=1 \Rightarrow \widehat{A}=\widehat{B}$
Vậy tam giác ABC đều (đpcm)