Trying to prove that if a random variable $T$ has Cauchy distribution with probability density function: $$f(x)= \frac{1}{\pi(1+x^2)}$$ then $X = \frac{1}{T}$ and $Y = \frac{2T}{1-T^2}$ are also Cauchy random variables.
Now, I realize that this specific form of PDF is called a Standard Cauchy distribution and if I am correct then $T$ is of the form:
$$ \frac{1}{e^{|t|}} $$
My main problem here is that I don't understand if i should use the characteristic function of $T$ to prove that $X$ and $Y$ are Cauchy or use it's PDF.
I apologize if this question is too simple, just this problem got me confused with the Cauchy distribution and I want to get the right picture of how to approach this type of questions, any hint greatly appreciated.
One way of obtaining a Cauchy random variable is as $T=\tan U$ where $U$ is a uniform random variable on the interval $(-\pi/2,\pi/2)$. Then $X=\cot U=\tan(\pi/2-U)$ and $Y=\tan 2U$. Now $\pi/2-U$ is uniform on $(0,\pi)$. Thinking of this modulo $\pi$, it's essentially the same as the distribution of $U$. Likewise $2U$ is uniform on $(-\pi,\pi)$. Again modulo $\pi$, that's essentially the same distribution as $U$.