Question

Consider triangle ABC in which A + B + C = π. Prove that

1.  tan A + tan B + tan C = tan A tan B tan C 

2.  tan (B/2) tan (C/2) + tan (C/2) tan (A/2) + tan (A/2) tan (B/2) = 1 

Answer 1

1.  We have A + B = π − C = 180° − C

2.  We have (A/2 + B/2) = p/2 − C/2 = 90° − C/2

⇒ tan(A/2 + B/2) = tan (p/2 − C/2) = cot(C/2)

Therefore, we get

tan(C/2) tan(A/2) + tan(B/2) tan(C/2) + tan(A/2) tan(B/2) = 1