Prove this Trigonomic Identity?

Question

tan(π/4+θ) - tan(π/4-θ) = 2tan2θ

I keep getting stuck on this one.
I was trying to use the double angle formula but messed up.

Answer 1

tan(π/4+θ) - tan(π/4-θ) =

[tan(π/4)+tanθ] / [ 1 - tan(π/4)·tanθ] - [tan(π/4)-tanθ] / [ 1 + tan(π/4)·tanθ] =

[1+tanθ] / [ 1 - tanθ] - [1-tanθ] / [ 1 +tanθ] =

[ (1+tanθ)^2 - (1-tanθ)^2] / [1-tan^2(θ)] =

4·tanθ / [1-tan^2(θ)] = 2 · 2·tanθ / [1-tan^2(θ)] =2·tan(2θ)

Saludos.

Answer 2

A = tan ( π/4 + θ ) = (1 + tan θ) / (1 - tan θ )
B = tan ( π/4 - θ ) = (1 - tan θ) / (1 + tan θ )

A - B is given by:-
4 tan θ / ( 1 - tan θ ) ( 1 + tan θ )
2 [ 2 tan θ / ( 1 - tan θ ) (1 + tan θ ) ]
2 tan 2θ

Answer 3

tan(π/4+θ)=tan[π/2-(π4-θ)]=
=cot(π/4-θ)=1/tan(π/4-θ)
so

tan(π/4+θ) - tan(π/4-θ)=
1/tan(π/4-θ) -tan(π/4-θ)=
=1-tan^2(π/4-θ)/tan(π/4-θ)=
=2[1-tan^2(π/4-θ)]/2tan(π/4-θ)=
=2/tan2(π/4-θ)=2/tan(π/2-2θ)=
=2/cot2θ=2tan2θ