Answer
The equation of the tangent line $(l)$ to $y$ at point $A(\pi,-1)$ would be $$(l): y=x-\pi-1$$
Work Step by Step
$$y=\cos x-\sin x$$
1) Find the derivative of $y$ $$y'=(\cos x-\sin x)'$$ $$y'=-\sin x-\cos x$$
2) Find $y'(\pi)$ $$y'(\pi)=-\sin\pi-\cos\pi$$ $$y'(\pi)=-0-(-1)=1$$
3) We know $y'(\pi)$, therefore the slope of the tangent line $(l)$ to $y$ at point $A(\pi,-1)$ is also known, since it equals $y'(\pi)$.
So the equation of the tangent line $(l)$ to $y$ at point $A(\pi,-1)$ would be $$(l): y-(-1)=y'(\pi)(x-\pi)$$ $$(l): y+1=1\times(x-\pi)$$ $$(l): y=x-\pi-1$$
