Answer
$y=x- \pi -1$ or, $y=x-(\pi +1)$
Work Step by Step
Given: $y= \cos x- \sin x$
$y'=- \sin x-\cos x$
The slope of tangent at $(\pi, -1)$, we have
$y'(\pi)=- \sin \pi-\cos \pi=0+1=1$
so we have $m=y'=1$
Formula to tangent line is : $y-y_1=m(x-x_1)$
$y-(-1)=1(x-\pi)$
This implies
$y+1=x-\pi$
or, $y=x- \pi -1$ or, $y=x-(\pi +1)$
