Ta có \(A+B+C=\pi\)
\(\Rightarrow A+B=\pi-C\)
\(\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)
\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\)
\(\Rightarrow tanA+tanB=-tanC\left(1-tanA.tanB\right)\)
\(\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\) ( đpcm )