How is $2(\ln \tan x + c)$ simplified to $A \tan^2 x$ where $A =2c$

Question

I'm trying to follow this reasoning:

$1/2 \ln(4 +y^2) = \ln(\tan x) + C$

$\ln(4 +y^2) = 2\ln(\tan x) + \ln A$ ( constant 2C = A)

$4 + y^2 = A \tan^2 x$

Answer 1

$\frac{1}{2}ln(4+ y^2)= ln(tan(x))+ C$ Multiply on both sides by 2: $ln(4+ y^2)= 2ln(tan(x))+ 2C$ Taking $A= e^{2C}$ (NOT A= 2C) we have $2C= ln(A)$ so $ln(4+ y^2)= ln(tan^2(x))+ ln(A)= ln(Atan^2(x))$

Now take the exponential of both sides: $e^{ln(4+ y^2)}= e^{ln(Atan^2(x))}$

$4+ y^2= Atan^2(x)$

I have used: 1) $2ln(x)= ln(x^2)$

2) $ln(a)+ ln(b)= ln(ab)$

3) $e^{ln(x)}= x$