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