<s>[INST] Show that $\det\left[\begin{smallmatrix}1&\cos a&\cos b\\ \cos a&1&\cos(a+b) \\ \cos b&\cos(a+b)&1 \end{smallmatrix}\right]=0$ I am unable to show - without expanding, by using determinant properties - that 
$$\det\begin{bmatrix}  1                       &\cos a               &\cos b\\
\cos a                     &1                &\cos(a+b) \\ 
\cos b                  &\cos(a+b)           &1
\end{bmatrix}=0$$
I am using trigonometric identities to solve this but I don't understand what would be the next step.
 [/INST] Here we go: directly
$$\begin{align}\begin{vmatrix}  1                       &\cos a               &\cos b\\
\cos a                     &1                &\cos(a+b) \\ 
\cos b                  &\cos(a+b)           &1
\end{vmatrix}&=\\&=
[1+2\cos a\cos b\cos(a+b)]-[\cos^2b+\cos^2a+\cos^2(a+b)]\\&=
1+\cos(a+b)\cdot [2\cos a\cos b-\cos (a+b)]-\cos^2a-\cos^2b\\&=
1+\cos(a+b)\cos(a-b)-\cos^2a-\cos^2b\\&=
1+\frac12[\cos(2a)+\cos(2b)]-\cos^2a-\cos^2b\\&=
1+\frac12[2\cos^2a-1+2\cos^2b-1]-\cos^2a-\cos^2b\\&=0\end{align}$$
Another method: triangulation
$$\begin{vmatrix}  1                       &\cos a               &\cos b\\
\cos a                     &1                &\cos(a+b) \\ 
\cos b                  &\cos(a+b)           &1
\end{vmatrix}\stackrel{(-\cos a)R_1+R_2\to R_2; \\(-\cos b)R_1+R_3\to R_3}{=}\\
\begin{vmatrix}  1                       &\cos a               &\cos b\\
0                     &1-\cos^2a                &\cos(a+b)-\cos a\cos b \\ 
0                  &\cos(a+b)-\cos a\cos b           &1-\cos^2b
\end{vmatrix}=\\
\begin{vmatrix}  1                       &\cos a               &\cos b\\
0                     &\sin^2a                &-\sin a\sin b \\ 
0                  &-\sin a\sin b           &\sin^2b
\end{vmatrix}\stackrel{\frac{\sin b}{\sin a}\cdot R_2+R_3\to R_3}{=}\\
\begin{vmatrix}  1                       &\cos a               &\cos b\\
0                     &\sin^2a                &-\sin a\sin b \\ 
0                  &0           &0
\end{vmatrix}=0.$$
</s>