Ta có:
$\dfrac{tanA}{tanC} = \dfrac{sin^2A}{sin^2C}$
$\Leftrightarrow \dfrac{sinA.cosC}{sinC.cosA} = \dfrac{sin^2A}{sin^2C}$
$\Leftrightarrow sinA.cosA = sinC.cosC$
$\Leftrightarrow sin2A = sin2C$
$\Leftrightarrow \left[\begin{array}{l}2A = 2C\\2A = 180^o - 2C\end{array}\right.$
Do $ABC$ không cân
nên $2A = 180^o - 2C$
$\Rightarrow 2A + 2C = 180^o$
$\Rightarrow A + C = 90^o$