Trong tam giác vuông ABH:
\(tanB=\dfrac{AH}{BH}\)
Trong tam giác vuông ACH:
\(tanC=\dfrac{AH}{CH}\)
\(\Rightarrow tanB+tanC=\dfrac{AH}{BH}+\dfrac{AH}{CH}=\dfrac{AH\left(BH+CH\right)}{BH.CH}=\dfrac{AH.BC}{BH.CH}=\dfrac{2AH^2}{BH.CH}\)
\(=2\left(\dfrac{AH}{BH}\right).\left(\dfrac{AH}{CH}\right)=2tanB.tanC\) (đpcm)