Answer: see proof below
Step-by-step explanation:
Use the following Sum Identity:
[tex]\tan (A + B) =\dfrac{\tan A+\tan B}{1-\tan A\cdot \tan B}[/tex]
Given: A + B + C = 180° → A + B + C = π
Proof LHS → RHS
Given: A + B + C = π
Multiply by 2: 2(A + B + C = π)
→ 2A + 2B + 2C = 2π
→ 2A + 2B = 2π - 2C
Apply tan: tan(2A + 2B = 2π - 2C)
→ tan (2A + 2B) = tan(2π - 2C)
→ tan (2A + 2B) = - tan 2C
[tex]\text{Sum Identity:}\qquad \qquad \dfrac{\tan 2A+\tan 2B}{1-\tan 2A\cdot \tan 2B}=-\tan 2C[/tex]
Simplify: tan 2A + tan 2B = -tan 2C (1 - tan 2A · tan 2B)
Distribute: tan 2A + tan 2B = -tan 2C + tan 2A · tan 2B · tan 2C
Add tan 2C: tan 2A + tan 2B + tan 2C = tan 2A · tan 2B · tan 2C
LHS = RHS is proven
