Answer
$-1-2x-\dfrac{5x^2}{2}+....$
Work Step by Step
The Maclaurin's series for $\cos x$ is as follows:
$1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+......=\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!}$
Now, we have the Maclaurin's series for $\dfrac{2}{1-x}$ is as follows:
$2(1+x+x^2+......+x^n+...)=2 \Sigma_{n=0}^{\infty} x^n$
Therefore,
$\cos x -\dfrac{2}{1-x}=1 -\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+......-[2(1+x+x^2+......+x^n+...)]=[\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!}]-[2 \Sigma_{n=0}^{\infty} x^n]$
or, $=-1-2x-\dfrac{5x^2}{2}+....$
