f(x) is  is even function

$f(x)\ is\ even \ function \ because\\ f(-x) = 1-\dfrac{2\pi}{x} ,-\pi \le-x\le0 \rightarrow 0\le x\le\pi \\ f(-x) = 1+\dfrac{2\pi}{x} ,0 \le-x\le\pi \rightarrow -\pi\le x\le0 \\ f(x) =f(-x) \therefore \ b_n = 0 $

$Let \ f(x) = a_0 + \displaystyle\sum _{n+1}^{\infty} a_n .\cos nx $

$\therefore \ a_0 = \int_{0}^{\pi} f(x).dx = \int_{0}^{\pi} (1-\dfrac{2\pi}{x}).dx\\ = \dfrac{1}{\pi} [x-\dfrac{x^2}{\pi}]_0^\pi = \dfrac{1}{\pi}(\pi -\pi) =0 $

$a_n = \dfrac{2}{\pi} \int_0^\pi .f(x) \ cosnx.dx\\ = \dfrac{2}{\pi} \int_0^\pi .(1-\dfrac{2x}{\pi}) \ cosnx.dx\\ $

integrating by parts

$\dfrac{2}{\pi} [ (1-\dfrac{2x}{\pi}) (\dfrac {sinnx}{n}) - (-\dfrac{-cosnx}{n^2}) ]_0^\pi$

$\dfrac{2}{\pi}.[( 0- \dfrac{2cosn\pi}{ n^2})-(\dfrac{-2}{\pi n^2})] \\ =\dfrac{4}{\pi^2 .n^2} [1-\ cosn\pi]\\ = 0 \ if \ n \ is\ even\\ =\dfrac{8}{\pi^2 n^2} if\ n \ is \ odd $

$\therefore f(x) = \displaystyle\sum _{n=1}^\infty a_n \ cosnx\\ =\dfrac{8}{\pi^2} \displaystyle\sum _{n=1}^\infty \dfrac{cosnx}{n^2}$

$\therefore f(x) = \dfrac{8}{\pi^2}[ \dfrac{cosx}{1^2} + \dfrac{cos3x}{3^2} + \dfrac{cos5x}{5^2} + \dfrac{cos7x}{7^2} + . ............. ]$