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

1 Answer
Steve M
Sep 20, 2017

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

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