Theo giả thiết ta có:
$\div tan \frac{\alpha}{2}, tan \frac{\beta}{2}, tan \frac{\gamma}{2} \Leftrightarrow tan \frac{\alpha}{2} + tan \frac{\gamma}{2}= 2 tan \frac{\beta}{2} \Leftrightarrow \dfrac{sin \frac{\alpha+\gamma}{2} }{cos \frac{\alpha}{2}cos \frac{\gamma}{2}} = 2 \dfrac{sin \frac{\beta}{2} }{cos \frac{\beta}{2} } $
Do $\alpha+\beta+\gamma=\pi \Rightarrow sin \frac{\alpha+\gamma}{2}=cos \frac{\beta}{2} $, bởi vậy
$\dfrac{sin \frac{\alpha+\gamma}{2} }{cos \frac{\alpha}{2}. cos \frac{\gamma}{2}}=2 \dfrac{sin \frac{\beta}{2} }{cos \frac{\beta}{2} } \Leftrightarrow 2 cos^2 \frac{\beta}{2} = 4 sin \frac{\beta}{2} cos \frac{\alpha}{2} cos \frac{\gamma}{2} $
$\Leftrightarrow 1+cos \beta = 2sin \frac{\beta}{2}\left ( cos \frac{\alpha+\gamma}{2}+ cos \frac{\alpha - \gamma}{2}\right ) $
$\Leftrightarrow 1+cos \beta = 2sin \frac{\beta}{2}. sin \frac{\beta}{2}+2 cos \frac{\alpha + \gamma}{2} cos \frac{\alpha-\gamma}{2} $
$\Leftrightarrow 1+cos \beta = 1- cos \beta+ cos \alpha+ cos \gamma$
$\Leftrightarrow 2cos \beta =cos \alpha + cos \gamma$
$\Leftrightarrow cos \alpha, cos \beta, cos \gamma$ lập thành cấp số cộng.