Can we say that $\det(A+B) = \det(A) + \det(B) +\operatorname{tr}(A) \operatorname{tr}(B) - \operatorname{tr}(AB)$? Let $A,B \in M_n$. Is this formula true?
$$\det(A+B) = \det(A) + \det(B) + \operatorname{tr}(A) \operatorname{tr}(B) - \operatorname{tr}(AB).$$

Here's an attempt to make the $2\times 2$ case seem just slightly horrible, while practicing to master the align* environment.
Let $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ and $B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}$.  
Then the left-hand side expands to eight terms:
\begin{align*}
\det (A+B) ={}& (a_{11}+b_{11})(a_{22}+b_{22}) - (a_{12}+b_{12})(a_{21}+b_{21}) \\
={}& \hphantom{(}a_{11}a_{22} - a_{12}a_{21} \\
&+ b_{11}b_{22} - b_{12}b_{21} \\
&+ a_{11}b_{22} - a_{21}b_{12} \\
&+ a_{22}b_{11} - a_{12}b_{21}
\end{align*}
The first two lines of the latter expression are respectively $\det A = a_{11}a_{22} - a_{12}a_{21}$ and $\det B=b_{11}b_{22} - b_{12}b_{21}$.  
That leaves only the expansion and cancellation for the trace expressions:
\begin{align*}
(\operatorname{tr} A)(\operatorname{tr} B) - \operatorname{tr}(AB) 
={}& (a_{11}+a_{22})(b_{11}+b_{22}) \\
&- (a_{11}b_{11} + a_{12}+b_{21} + a_{21}+b_{12} + a_{22}+b_{22}) \\
={}& a_{11}b_{22} - a_{21}b_{12} \\
&+ a_{22}b_{11} - a_{12}b_{21}
\end{align*}
Thus in the $2\times 2$ cases the identity is valid:
$$ \det A+B = \det A + \det B + (\operatorname{tr} A)(\operatorname{tr} B) - \operatorname{tr}(AB) $$
