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.
 A: This fact is useful in thinking about your problem: If $U,V\sim N(0,1)$ are independent standard normals, their ratio $X=U/V$ has the standard Cauchy distribution.  Then $Y=(1-X)/(1+X) = (V-U)/(V+U) = S/T$ where $S=(V-U)/\sqrt 2$ and $T=(V+U)/\sqrt 2$.  But $S$ and $T$ are also independent standard normals, so $Y$ has the same distribution as $X$.
Equivalently, one can represent the iid variables $U,V\sim N(0,1)$ as $U=R\cos\Theta, V=R\sin\Theta$, where $R,\Theta$ are independent, with $R$ Rayleigh distributed and $\Theta$ uniform on $[0,2\pi)$, so $X=\tan\Theta$ and $Y=(1-X)/(1+X)=\tan(\pi/4-\Theta)$.  Since $\Theta$ is uniform, so is $\pi/4-\Theta$, so $Y$ has the same distribution as $X$.
A: 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.
A: $$P((1-X)/(X+1)\leq t)=\begin{cases}
P(\{X\leq -1\}\cup\{X\geq (1-t)/(1+t)\})&t>-1\\
P(\{(1-t)/(1+t)\leq X\leq -1\}& t<-1
\end{cases}$$
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 $t<-1$
$$P((1-X)/(X+1)\leq t)=\frac{1}{2}+\frac{1}{\pi}\arctan(-1)-\frac{1}{2}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}=\\-\frac{1}{4}-\frac{1}{\pi}\arctan\frac{1-t}{1+t}$$
A: Remark (Theorem of transformation for one random variable): Let be $X$ a continuous r.v. with pdf given by $f_X$ and let be Y=g(X) with g a diffeomorphism, then the pdf $f_Y$ is given by:

$f_Y(y)=f_X(g^{-1}(y))\cdot|{{d\over{dy}}g^{-1}(y)}|$.

If $Y=g(X)={{1+X}\over{1-X}}$ then $X=g^{-1}(Y)={{1-Y}\over{1+Y}}$. We find that ${dx\over{dy}}=-{{2\over{1+y^2}}}$. Now:

$f_Y(y)=f_X({{1-y}\over{1+y}})\cdot{2\over{1+y^2}}$.

I'll let you do the math, then you get:

$f_Y(y)={1\over{\pi{(1+y^2)}}}$.

So $Y\sim{Cauchy(0,1)}$.
