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.

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 $X$.
