Answer
(a) $z=x+1$
(b) $x=1-z$, $y=0$
Work Step by Step
$z=e^xcosy$ and $(0,0,1)$
(a) Equation for a tangent plane is given as:
$z-1=1(x-0)+0(y-0)$
$z-1=x$
$z=x+1$
(b) $x-z+1=0$
Equation of normal line is:
$\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-1}{-1}$
$x=1-z$, $y=0$
