<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] Let $g(x) = (1-x)/(1+x)$.  Then $g$ is self-inverse; i.e., $g^{-1} = g$, but it is not everywhere monotone.  Thus, consider $\Pr[g(X) \le y]$ for the case $y \le -1$ versus $y > -1$ separately.  In the first case, $g$ is monotone decreasing on $X \in (-\infty, -1)$ and we have $$\Pr[g(X) \le y] = \Pr[g^{-1}(y) \le X < -1] = \Pr[g(y) \le X < -1],$$ with no issues (the inequality reverses because $g$ is order-reversing).  You can check this with a numerical example; e.g., for $y = -2$, $$\frac{1-x}{1+x} \le -2 \iff -3 \le x \le -1.$$
However, for $y > -1$, we see that the inequality is compound:  $$\begin{align}
\Pr[g(X) \le y] &= \Pr[-\infty < X < -1] + \Pr[g^{-1}(y) \le X < \infty] \\
&= 1 - \Pr[-1 < X \le g(y)]. \end{align}$$
Hence we have $$F_Y(y) = \int_{x=g(y)}^{-1} f_X(x) \, dx$$ when $y \le -1$ and $$F_Y(y) = \int_{x = -\infty}^{-1} f_X(x) \, dx + \int_{x=g(y)}^\infty f_X(x) \, dx = 1 - \int_{x=-1}^{g(y)} f_X(x) \, dx$$ when $y > -1$.  I leave the rest of the computation as an exercise.
</s>