How tan(π/4+θ) = (1+tanθ)/(1-tanθ) ?

Derivation of

tan(π/4+θ) = (1+tanθ)/(1-tanθ)

As,  tan(A+B) = (tanA - tanB)/(1- tanA tanB)

       👉Derivation of tan(A+B) = (tanA + tanB)/(1- tanA tanB)

{Put A = π/4 and B= θ}

tan(π/4+θ) =

[tan(π/4) + tanθ] /[1- tan(π/4)tanθ]

As tan(π/4) = 1.......(i.e. tan 45° = 1)

Hence, 

tan(π/4+θ)= (1+tanθ)/(1-tanθ)