Question

Please someone help me to prove this. ​

Answer 1

Hope this can help.

Answer 2

Answer:  see proof below

Step-by-step explanation:

Use the following Sum Identity:

$\tan (A + B) =\dfrac{\tan A+\tan B}{1-\tan A\cdot \tan B}$

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

$\text{Sum Identity:}\qquad \qquad \dfrac{\tan 2A+\tan 2B}{1-\tan 2A\cdot \tan 2B}=-\tan 2C$    

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