Given a function $f(z) = \cos\left(\dfrac{z+3}{z+2}\right)$, we are required to find its Laurent expansion at $z = -2$.
To cope with such problem, we start with a change of variable. We start with $w = z + 2$, then we seek to expand it along $\cos(1+\frac{1}{w})$, which is
$$ 1 - \frac{1}{2!}\left(1+\frac{1}{w}\right)^2 + \frac{1}{4!}\left(1+\frac{1}{w}\right)^4 \pm \ldots $$
Then I think of expanding the formula to get the coefficient of $\frac{1}{w}$, $\frac{1}{w^2}$,... separately. But it just seems tedious, and I don't feel like techniques relating to Laurent Series are included in this brutal force method...
The suggestion in the comments is very useful. We have by the cosine addition formula \begin{align} \cos\left(1+\frac{1}{w}\right)&=\cos(1)\cos\left(\frac{1}{w}\right)-\sin(1)\sin\left(\frac{1}{w}\right)\\ &=\cos(1)\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\left(\frac{1}{w}\right)^{2n}- \sin(1)\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}\left(\frac{1}{w}\right)^{2m+1}. \end{align} This gives the decomposition into the even and odd powers.