Question

Prove that `ln( (1+tan(x/2)) / (1-tan(x/2)) )=ln(sec(x)+tan(x))`?

Answer 1

We want to show that:

`ln( (1+tan(x/2)) / (1-tan(x/2)) )=ln(sec(x)+tan(x))`

Due to the monotonicity of the logarithmic function, this is the same as showing:

`(1+tan(x/2)) / (1-tan(x/2)) = sec(x)+tan(x)` ..... [A]

For simplicity, put `t=tan(x/2)`
We will need the tangent half-angle formula:

`sin A -= (2tan (A/2))/(1+tan^2(A/2)) => sinx = (2t)/(1+t^2)`

`cos A -= (1-tan^2(A/2))/(1+tan^2(A/2)) => cos x =(1-t^2)/(1+t^2)`

`tan A -= (2tan (A/2))/(1-tan^2(A/2)) => tan x = (2t)/(1-t^2)`

Then, if we concentrate on the LHS of [A], we have:

`LHS = (1+t) / (1-t)`

`\ \ \ \ \ \ \ \ = (1+t) / (1-t) * (1+t)/(1+t)`

`\ \ \ \ \ \ \ \ = (1+2t+t^2) / (1-t^2)`

`\ \ \ \ \ \ \ \ = ((1+t^2) + (2t)) / (1-t^2)`

`\ \ \ \ \ \ \ \ = (1+t^2)/ (1-t^2) + (2t)/ (1-t^2)`

`\ \ \ \ \ \ \ \ = 1/cosx + tanx`

`\ \ \ \ \ \ \ \ = secx + tanx` QED