Find the density of $Y = a/(1 + X^{2})$, where $X$ has the Cauchy distribution. Find the density of $Y = a/(1 + X^{2})$, where $X$ has the Cauchy distribution.
MY SOLUTION
To begin with, let us remember that the Cauchy probability density function is given by
\begin{align*}
f_{X}(x) = \frac{1}{\pi(1+x^{2})}\quad\text{for}\quad x\in\textbf{R}
\end{align*}
Consequently, the sought distribution is described as
\begin{align*}
F_{Y}(Y\leq y) &= \textbf{P}(Y\leq y) = \textbf{P}\left(\frac{a}{1+X^{2}} \leq y\right) = \textbf{P}\left(1+X^{2}\geq \frac{a}{y}\right)\\\\
& = \textbf{P}\left(|X|\geq \sqrt{\frac{a-y}{y}}\right) = \textbf{P}\left(X\geq \sqrt{\frac{a-y}{y}}\right) + \textbf{P}\left(X \leq -\sqrt{\frac{a-y}{y}}\right)\\\\
& = 1 - F_{X}\left(\sqrt{\frac{a-y}{y}}\right) + F_{X}\left(-\sqrt{\frac{a-y}{y}}\right)
\end{align*}
where $y\in(0,a]$. Finally, we have
\begin{align*}
F_{X}(x) = \int_{-\infty}^{x}f_{X}(u)\mathrm{d}u = \int_{-\infty}^{x}\frac{\mathrm{d}x}{\pi(1+x^{2})} = \frac{\arctan(x)}{\pi} + \frac{1}{2}\end{align*}
I have two questions. Firstly, I would like to know if this approach is correct. Secondly, I would like to know if there is another way to solve this problem. Thanks in advance!

Your approach is fine. 
Perhaps you are aware of this result (very useful to generate a Cauchy variable): if $Z \sim U(-\frac{\pi}{2},\frac{\pi}{2})$ then $X = \tan(Z)$ follows a canonical Cauchy distribution.
Now, we have $Y = \frac{a}{1+X^2}= a \cos^2(Z)$ which is slightly easier. 
Then, assuming $a>0$ $$\begin{align}
f_Y(y) &= 2 \frac{f_Z(z)}{|2 a \cos(z) \sin(z)|} \\
&=\frac{1}{a \pi} \frac{1}{|\cos^2(z)\tan(z)|}\\
&=\frac{1}{a \pi} \frac{1}{|\frac{y}{a}x|}\\
&=\frac{1}{ \pi} \frac{1}{y\sqrt{a/y-1}} \\
&=\frac{1}{ \pi} \frac{1}{\sqrt{y(a-y)}}
\end{align}
$$
for $0<y<a$
