Find an exact normal on a trigonometric curve

Both copies print the final answer without working. Independently, differentiating in radians gives $\cos x-\sin y\,dy/dx=0$, so $dy/dx=\cos x/\sin y$. At the given point, both numerator and denominator equal $\sqrt3/2$, giving slope 1.

The normal has slope $-1$. Through $(\pi/6,\pi/3)$ it is $y-\pi/3=-(x-\pi/6)$, hence $y=-x+\pi/2$, matching both printed answers.