Answer
$(\frac{\pi}{4}, 1)$ and $(-\frac{\pi}{4}, -1)$.
See graph.
Work Step by Step
Step 1. Given $y=tan(x)$, we have $y'=sec^2x$ which gives the slope of tangent lines to the curve.
Step 2. Given the line $y=-x/2$, we can find its slope as $m=-1/2$. For the tangent line to be normal to this line, we have $my'=-1, y'=2$ which gives $sec^2x=2$ and $cos(x)=\pm\frac{\sqrt 2}{2}$
Step 3. Within $-\pi/2\lt x\lt \pi/2$, the solution to the above equation gives $x=\pm\frac{\pi}{4}$, which gives $y=\pm1$
Step 4. The points for the answer to this question are $(\frac{\pi}{4}, 1)$ and $(-\frac{\pi}{4}, -1)$.
Step 5. The equations for the normals are $y-1=-\frac{1}{2}(x-\frac{\pi}{4})$ or $y=-\frac{1}{2}x+\frac{\pi}{8}+1$ and $y+1=-\frac{1}{2}(x+\frac{\pi}{4})$ or $y=-\frac{1}{2}x-\frac{\pi}{8}-1$
Step 6. See graph.
