<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] To flesh out my comment from earlier,$$\Delta:=\left|\begin{array}{ccc}
1 & \cos a & \cos b\\
\cos a & 1 & \cos\left(a+b\right)\\
\cos b & \cos\left(a+b\right) & 1
\end{array}\right|=\left|\begin{array}{ccc}
1 & \cos a & \cos b\\
\cos a & 1 & \cos\left(a+b\right)\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|=\left|\begin{array}{ccc}
1 & \cos a & 0\\
\cos a & 1 & -\sin a\sin b\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|.$$We may as well also simplify the second row/column:
$$\Delta=\left|\begin{array}{ccc}
1 & 0 & 0\\
\cos a & \sin^{2}a & -\sin a\sin b\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|=\left|\begin{array}{ccc}
1 & 0 & 0\\
0 & \sin^{2}a & -\sin a\sin b\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|.$$Well, now we just have an easy $2\times 2$ determinant.
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