Answer
$y=2x- \pi$
Work Step by Step
Given: $y= x + \tan x$
$y'=1+ \sec^2 x$
The slope of tangent at $(\pi, \pi)$, we have
$y'(\pi)=1+ (-1)^2=1+1=2$
so we have $m=y'=1$
Formula to tangent line is : $y-y_1=m(x-x_1)$
$y-\pi=2(x-\pi)$
This implies
$y=2(x-\pi)+-\pi$
or, $y=2x- \pi$
