From the text is says: The Cauchy cdf is $$F(x) = \frac{1}{2}+\frac{1}{\pi}\arctan x, \qquad \infty < x < \infty$$ We are then asked to
(a) Find the density function, and
(b) Find $x$ such that $P(X > x) = .1$
My answer to (a) was to differentiate the cdf which got me $$f_X(x) = \frac{1}{\pi(x^2+1)}.$$
For part (b) I'm struggling. So far my solution is as follows $$\begin{align*} P(X > x) & = 1 - P(X \leq x)\\ 0.1&=1-\frac{1}{\pi}\int_{-\infty}^x \frac{1}{t^2+1}dt \\0.9\pi&=\lim_{b\to-\infty}\left[\int_b^x\frac{1}{t^2+1}dt\right]\\ 0.9\pi&=\lim_{b \to -\infty}\left[\arctan x - \arctan b\right]\\\pi1.4&= \arctan x \\ \tan(\pi1.4)&=x.\end{align*}$$ Now, $\tan(1.4\pi) \approx 0.0769$ and so $x$ should be that, however the answer in the text is $x=3.08.$ I'm not sure where I went wrong.
Notice that $$ \arctan{b} \to \color{red}{-}0.5\pi, \quad b\to-\infty, $$ so that $$ 0.9\pi = \lim_{b\to-\infty} (\arctan x - \arctan b) = \arctan x - (-0.5\pi) \quad \iff \quad 0.4\pi = \arctan x. $$ And indeed $x = \tan(0.4\pi) \approx 3.08$.