Answer
$y = \pi $
Work Step by Step
$$\eqalign{
& y = x + \sin x,{\text{ }}\left( {\pi ,\pi } \right) \cr
& {\text{Differentiate }}y{\text{ to calculate the slope at the point }}\left( {\pi ,\pi } \right) \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x + \sin x} \right] \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ x \right] + \frac{d}{{dx}}\left[ {\sin x} \right] \cr
& \frac{{dy}}{{dx}} = 1 + \cos x \cr
& {\text{Find }}m{\text{ at }}\left( {\pi ,\pi } \right) \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = \pi }} = 1 + \cos \left( \pi \right) \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = \pi }} = 0 \cr
& \cr
& {\text{Use the Point}} - {\text{Slope Form of the Equation of a Line}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \underbrace {\left( {\pi ,\pi } \right)}_{\left( {{x_1},{y_1}} \right)} \to x = \pi {\text{ and }}{y_1} = \pi \cr
& {\text{Therefore}} \cr
& y - \pi = 0\left( {x - \pi } \right) \cr
& {\text{Simplify}} \cr
& y = \pi \cr} $$
