Answer
$$ - 2x + z = 2$$
Work Step by Step
$$\eqalign{
& z = {e^x}\left( {\sin y + 1} \right),{\text{ }}\left( {0,\frac{\pi }{2},2} \right) \cr
& z - {e^x}\left( {\sin y + 1} \right) = 0 \cr
& {\text{Considering }} \cr
& F\left( {x,y,z} \right) = z - {e^x}\left( {\sin y + 1} \right) \cr
& {\text{Calculate the partial derivatives }} \cr
& {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {z - {e^x}\left( {\sin y + 1} \right)} \right] = - \left( {\sin y + 1} \right){e^x} \cr
& {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {z - {e^x}\left( {\sin y + 1} \right)} \right] = - {e^x}\cos y \cr
& {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {z - {e^x}\left( {\sin y + 1} \right)} \right] = 1 \cr
& {\text{At the point }}\left( {0,\frac{\pi }{2},2} \right){\text{ the partial derivatives are}} \cr
& {F_x}\left( {0,\frac{\pi }{2},2} \right) = - \left( {\sin \left( {\frac{\pi }{2}} \right) + 1} \right){e^0} = - 2 \cr
& {F_y}\left( {0,\frac{\pi }{2},2} \right) = - {e^0}\cos \left( {\frac{\pi }{2}} \right) = 0 \cr
& {F_z}\left( {0,\frac{\pi }{2},2} \right) = 1 \cr
& {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr
& {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr
& + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr
& {\text{Substituting}} \cr
& - 2\left( {x - 0} \right) + 0\left( {y - \frac{\pi }{2}} \right) + \left( {z - 2} \right) = 0 \cr
& {\text{Simplifying}} \cr
& - 2x + z - 2 = 0 \cr
& - 2x + z = 2 \cr} $$