Question: ￼$$\ln(\tan(x) + \sec(x))=\frac{\ln (1+\sin x )-\ln(1-\sin x )}{2}$$

Is this true or false? I thought the right side looked like a sum of an even and an odd function but I ended up with $$\ln(\tan(x) + \sec(x))=\ln(\tan(x) + \sec(x))$$ which is true, but it's not what I was trying to prove.

Answer: 
$2\ln (\tan x+\sec x)=\ln \left(\dfrac{(1+\sin x)^2}{\cos^2 x}\right)=\ln \left(\dfrac{(1+\sin x)^2}{1-\sin^2 x}\right)=\ln \left(\dfrac{1+\sin x}{1-\sin x}\right)$
Let $s = \sin x, c = \cos x$.

The LHS is $\displaystyle \ln (\frac{s}{c} + \frac{1}{c}) = \ln (\frac{s+1}{c}) = \ln(\frac{1+s}{\sqrt{1-s^2}}) = \ln (\frac{1+s}{\sqrt{(1+s)(1-s)}}) \\ =\ln(\frac{\sqrt{1+s}}{\sqrt{1-s}}) = \ln(\sqrt{1+s}) - \ln(\sqrt{1-s}) = \frac{1}{2} [\ln(1+s) - \ln(1-s)]$