The determinant$\left| {\begin{array}{llllllllllllllllllll}{\sin A}&{\cos A}&{\sin A + \cos B}\\{\sin B}&{\cos A}&{\sin B + \cos B}\\{\sin C}&{\cos A}&{\sin C + \cos B}\end{array}} \right|$ is equal to zero.

Correct Answer True

Since, $\left| {\begin{array}{llllllllllllllllllll}{\sin A}&{\cos A}&{\sin A + \cos B}\\{\sin B}&{\cos A}&{\sin B + \cos B}\\{\sin C}&{\cos A}&{\sin C + \cos B}\end{array}} \right|$

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

$= 0 + \left| {\begin{array}{llllllllllllllllllll}{\sin A}&{\cos A}&{\cos B}\\{\sin B}&{\cos A}&{\cos B}\\{\sin C}&{\cos A}&{\cos B}\end{array}} \right|$

[since, in first determinant ${C_1}$ and ${C_3}$

are identicals] $= \cos A \cdot \cos B\left| {\begin{array}{llllllllllllllllllll}{\sin A}&1&1\\{\sin B}&1&1\\{\sin C}&1&1\end{array}} \right|$

[taking ${\rm{cos }}A$ common from ${C_2}$

and $\cos B$ common from ${C_3}$]
$= 0$

[since, ${C_2}$

and ${C_3}$ are identicals]