Please help me as soon as possible ...​

Answer:  see proof below

Step-by-step explanation:

Use the following identities

tan (A + B) = (tan A + tan B)/(1 - tan A · tan B) -->  

tan 60° = √3

tan 120° = -√3

tan 3A = (3tanA - tan³A)/(1 - 3 tan²A)  -->

Proof LHS → RHS

Given:                  tan Ф + tan(60° + Ф) + tan(120° + Ф)

Sum Difference: tan Ф + (tan 60° + tanФ)/(1-tan60°·tanФ) + (tan 120° + tanФ)/(1-tan120°·tanФ)

              (latex)    

Substitute:         tan Ф + (√3 + tanФ)/(1-√3·tanФ) + (-√3 + tanФ)/(1+√3°·tanФ)

               (latex)  

Common Denominator: [tan Ф(1-3tan²Ф)+8tanФ]\(1-3tan²Ф)

                (latex)  

Distribute:             (tan Ф - 3tan³Ф + 8Ф)\(1 - 3 tan²Ф)

               (latex)    

Simplify:               (9Ф - 3tan³Ф)\(1 - 3 tan²Ф)

                          3(3Ф - tan³Ф)\(1 - 3 tan²Ф)

               (latex)  

Triple Angle Identity:   3 tan 3Ф

3 tan 3Ф = 3 tan 3Ф  

Your question has been heard loud and clear.

tanθ+tan(60∘+θ)+tan(120∘+θ)

=tanθ+3–√+tanθ1−3–√tanθ+−3–√+tanθ1+3–√tanθ

[tanθ(1−3tan2θ)+(3–√+tanθ)(1+3–√tanθ)

=+(−3–√+tanθ)(1−3–√tanθ)1−3tan2θ

=9tanθ−3tan3θ1−3tan2θ=3tan3θ

Thank you.