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# True or False: The determinant $\begin{vmatrix}\sin A & \cos A & \sin A+\cos B\\\sin B & \cos A & \sin B+\cos B\\\sin C & \cos A & \sin C+\cos B\end{vmatrix}$ is equal to zero

$\begin{array}{1 1} True \\ False \end{array}$

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Toolbox:
• If some or all elements of a row(or column) of determinant are expressed as sum of two (or more)terms,then the determinant can be expressed as sum of two (or more)determinants.
given:$\begin{vmatrix}\sin A & \cos A & \sin A+\cos B\\\sin B & \cos A & \sin B+\cos B\\\sin C & \cos A & \sin C+\cos B\end{vmatrix}$
As given by the information in the toolbox,let us split the determinant as
$\begin{vmatrix}\sin A & \cos A & \sin A\\\sin B & \cos A & \sin B\\\sin C & \cos A & \sin C\end{vmatrix}+\begin{vmatrix}\sin A & \cos A & \cos B\\\sin B & \cos A &\cos B\\\sin C & \cos A & \cos B\end{vmatrix}$
We know that only two rows(or columns)in a determinant are identical,then the value of the determinant is zero.
Let $\Delta=\Delta_a+\Delta_b$
$\Delta_a=\begin{vmatrix}sin A& cos A &sin A\\sin B& cos A &sin B\\sin C& cos A &sin C\end{vmatrix}$
Here $C_1$ and $C_3$ are identical.Hence the value is zero.
$\Delta_a=0.$
$\Delta_b=\begin{vmatrix}sin A& cos A &cos B\\sin B& cos A &cos B\\sin C& cos A &cos B\end{vmatrix}$
Taking cos A as the common factor from$C_1$ and cos B as a common factor from $C_3$ we get,
$cos A cos B\begin{vmatrix}sin A & 1& 1\\sin B & 1& 1\\sin C & 1& 1\end{vmatrix}$
Therefore $\Delta_b=0$.
Now $C_2$ and $C_3$ are identical,hence its value is zero.
$\Delta=0+0=0.$
Hence True.
answered Mar 27, 2013

+1 vote