Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 142: 44

$y=-x+\frac{\pi}{2}$ See graph.

Step 1. Given $y=cot(x)$, we have $y'=-csc^2x$ Step 2. For a tangent line parallel to the line $y=-x$, let $y'=-1$, we have $-csc^2x=-1$ and $csc(x)=\pm1$ thus $x=\pi/2$ in $(0,\pi)$ Step 3. At $x=\pi/2$, $y=cot(\pi/2)=0$, and the tangent line equation is $y=-(x-\pi/2)$ or $y=-x+\frac{\pi}{2}$ Step 4. See graph.