Answer
(a) $y=-\frac{\pi}{2}x+\pi$
(b) $y=\frac{2}{\pi}x-\frac{2}{\pi}+\frac{\pi}{2}$
Work Step by Step
To verify that the given point is on the curve, plug-in the coordinates to the left side of the equation: $LHS=2(1)(\pi/2)+\pi sin(\pi/2)=\pi+\pi=2\pi=RHS$
(a) Given the equation $2xy+\pi sin(y)=2\pi$, differentiate both sides with respect to $x$: $2y+2xy'+\pi cos(y)y'=0$, which gives $y'=-\frac{2y}{2x+\pi cos(y)}$
Thus the slope of the tangent line at the given point is $m=y'=-\frac{2(\pi/2)}{2+\pi cos(\pi/2)}=-\frac{\pi}{2}$ and the equation for the tangent line is $y-\frac{\pi}{2}=-\frac{\pi}{2}(x-1)$ or $y=-\frac{\pi}{2}x+\pi$
(b) The slope of the line normal to the tangent line is $n=-\frac{1}{m}=\frac{2}{\pi}$, and the line equation is $y-\frac{\pi}{2}=\frac{2}{\pi}(x-1)$ or $y=\frac{2}{\pi}x-\frac{2}{\pi}+\frac{\pi}{2}$
