Question

Chứng minh với mọi tam giác không vuông ABC có:

a, tan A + tan B + tan C = tan A . tan B . tan C

b, tan 2A + tan 2B + tan 2C = tan 2A . tan 2B . tan 2C ( A, B, C ≠ \(\frac{\text{π}}{4}\) )

Answer 1

\(A+B+C=180^0\Rightarrow tan\left(A+B\right)=-tanC\)

\(\Rightarrow\frac{tanA+tanB}{1-tanA.tanB}=-tanC\Leftrightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)

\(\Leftrightarrow tanA+tanB+tanC=tanA.tanB.tanC\)

\(2A+2B+2C=360^0\Rightarrow tan\left(2A+2B\right)=-tan2C\)

\(\Leftrightarrow\frac{tan2A+tan2B}{1-tan2A.tan2B}=-tan2C\)

\(\Leftrightarrow tan2A+tan2B+tan2C=tan2A.tan2B.tan2C\)