<s>[INST] Suppose X$\sim$ Cauchy(0,1). Then what will be the distribution of $\frac{1-X}{1+X}$? In order to find distribution of $\frac{1-X}{1+X}$ below approach I followed,
Let,
\begin{align}
    Y = \frac{1-X}{1+X}
\end{align}
Then, cdf of Y is
\begin{align}
    F_{Y}(y) = P(Y \leq y)
\end{align}
\begin{align}
    = P\left(\frac{1-X}{1+X} \leq y\right)
\end{align}
\begin{align}
    = 1 - P\left(X < \frac{1-y}{1+y}\right)
\end{align}
\begin{align}
    = 1 -  \int_{-\infty}^{\frac{1-y}{1+y}} f(x) \,dx
\end{align}
\begin{align}
    = 1 -  \int_{-\infty}^{\frac{1-y}{1+y}} \frac{1}{\pi}\cdot \frac{1}{1+x^2} \,dx
\end{align}
\begin{align}
    = 1 -  \frac{1}{\pi}\cdot \left[tan^{-1}x\right]_{-\infty}^{\frac{1-y}{1+y}}
\end{align}
\begin{align}
    F_{Y}(y) = \frac{1}{\pi}\cdot \left[-\frac{\pi}{2}+tan^{-1}\left({\frac{1-y}{1+y}}\right)\right]
      = \frac{1}{2} -\frac{1}{\pi}.tan^{-1}\left({\frac{1-y}{1+y}}\right)
\end{align}
and then
\begin{align}
    f_{Y}(y) = \frac{d F_{Y}(y)}{dy} = \frac{1}{\pi}\cdot \frac{1}{1+y^2}
\end{align}
But I have a little confusion here how to find range of Y from X? And CDF of Y is doesn't looks like cdf of a cauchy distribution.
 [/INST] $$P((1-X)/(X+1)\leq t)=\begin{cases}
P(\{X\leq -1\}\cup\{X\geq (1-t)/(1+t)\})&t>-1\\
P(\{(1-t)/(1+t)\leq X\leq -1\}& t<-1
\end{cases}$$
So for $t>-1$
$$P((1-X)/(X+1)\leq t)=\frac{1}{2}+\frac{1}{\pi}\arctan(-1)+\frac{1}{2}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}=\\\frac{3}{4}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}$$
while for $t<-1$
$$P((1-X)/(X+1)\leq t)=\frac{1}{2}+\frac{1}{\pi}\arctan(-1)-\frac{1}{2}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}=\\-\frac{1}{4}-\frac{1}{\pi}\arctan\frac{1-t}{1+t}$$
</s>