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: 
We'll start from RHS
$$\begin{align}
\frac12\Big[\ln(1+\sin x )-\ln(1-\sin x)\Big]
&=\frac12\left[\ln\left(\frac{1+\sin x}{1-\sin x} \right)\right]\\
&=\frac12\left[\ln\left(\frac{(1+\sin x)^2}{1-\sin^2 x} \right)\right]\\
&=\frac12\left[\ln\left(\frac{(1+\sin x)^2}{\cos^2 x} \right)\right]\\
&=\ln\left(\frac{ 1+\sin x }{\cos x} \right)\\
&=\ln\left(\frac1{\cos x}+\frac{\sin x}{\cos x} \right)\\
\frac12\Big[\ln(1+\sin x )-\ln(1-\sin x)\Big]
&=\ln\left(\sec x+\tan x \right)\\
\end{align}$$

So, It's true!
$2\ln (\tan x+\sec x)=\ln \left(\dfrac{(1+\sin x)^2}{\cos^2 x}\right)=\ln \left(\dfrac{(1+\sin x)^2}{1-\sin^2 x}\right)=\ln \left(\dfrac{1+\sin x}{1-\sin x}\right)$