The value of the determinant

Let $\Delta=\left|\begin{array}{ccc}\sin A & \cos A & \sin A+\cos B \\ \sin B & \cos A & \sin B+\cos B \\ \sin C & \cos A & \sin C+\cos B\end{array}\right|$

$\Delta=\left|\begin{array}{lll}\sin A & \cos A & \sin A+\cos B \\ \sin B & \cos A & \sin B+\cos B \\ \sin C & \cos A & \sin C+\cos B\end{array}\right|$

$=\left|\begin{array}{lll}\sin A & \cos A & \sin A \\ \sin B & \cos A & \sin B \\ \sin C & \cos A & \sin C\end{array}\right|+\left|\begin{array}{ccc}\sin A & \cos A & \cos B \\ \sin B & \cos A & \cos B \\ \sin C & \cos A & \cos B\end{array}\right|$

$=0+\left|\begin{array}{lll}\sin A & \cos A & \cos B \\ \sin B & \cos A & \cos B \\ \sin C & \cos A & \cos B\end{array}\right|$                 ( $\because$ if any two columns are identical then the value of determinant is zero)

$=\left|\begin{array}{lll}\sin A & \cos A & \cos B \\ \sin B & \cos A & \cos B \\ \sin C & \cos A & \cos B\end{array}\right|$

Taking out $\cos A$ and $\cos B$ common from $C_{2}$ and $C_{3}$, respectively

$=\cos A \cos B\left|\begin{array}{lll}\sin A & 1 & 1 \\ \sin B & 1 & 1 \\ \sin C & 1 & 1\end{array}\right|$

$=\cos A \cos B(0)$        ( $\because$ if any two columns are identical then the value of determinant is zero)

$=0$

Hence, the value of the determinant $\left|\begin{array}{lll}\sin A & \cos A & \sin A+\cos B \\ \sin B & \cos A & \sin B+\cos B \\ \sin C & \cos A & \sin C+\cos B\end{array}\right|$ is $\underline{0}$.