Let $A,$ $B,$ $C$ be the angles of a non-right triangle.  Compute
\[\begin{vmatrix} \tan A & 1  & 1 \\ 1 & \tan B & 1 \\ 1 & 1 & \tan C \end{vmatrix}.\]
I remember that in a triangle, the sum of the angles is 180. Yes, so $A+B+C=180$. Right. So we can rewrite $\tan C=\tan(180^\circ-A-B)$. Also, $\tan(180^\circ-A-B)=-\tan(A+B)$. And we know that $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. So we can substitute for $\tan C$ in the determinant. Let's call the determinant $D$. Then $D=\begin{vmatrix} \tan A & 1  & 1 \\ 1 & \tan B & 1 \\ 1 & 1 & -\frac{\tan A+\tan B}{1-\tan A\tan B} \end{vmatrix}$. Now we expand. So our determinant is $\left(-\frac{\tan A+\tan B}{1-\tan A\tan B}\right)\tan B\tan C+2-\left(-\frac{\tan A+\tan B}{1-\tan A\tan B}\right)-\tan B-\tan C$. Right. We can simplify this to $2$.

