Question

The value of the determinant $\left|\begin{array}{ccc}\sin A& \cos A& \sin A+\cos B\\ \sin B& \cos A& \sin B+\cos B\\ \sin C& \cos A& \sin C+\cos B\end{array}\right|$ is ________________.

Answer 1

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


$$\begin{gathered} ∆=\left|\begin{array}{ccc}\sin A& \cos A& \sin A+\cos B\\ \sin B& \cos A& \sin B+\cos B\\ \sin C& \cos A& \sin C+\cos B\end{array}\right| \\ =\left|\begin{array}{ccc}\sin A& \cos A& \sin A\\ \sin B& \cos A& \sin B\\ \sin C& \cos A& \sin C\end{array}\right|+\left|\begin{array}{ccc}\sin A& \cos A& \cos B\\ \sin B& \cos A& \cos B\\ \sin C& \cos A& \cos B\end{array}\right| \\ =0+\left|\begin{array}{ccc}\sin A& \cos A& \cos B\\ \sin B& \cos A& \cos B\\ \sin C& \cos A& \cos B\end{array}\right|\left(\because \mathrm{if}\mathrm{any}\mathrm{two}\mathrm{columns}\mathrm{are}\mathrm{identical}\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{determinant}\mathrm{is}\mathrm{zero}\right) \\ =\left|\begin{array}{ccc}\sin A& \cos A& \cos B\\ \sin B& \cos A& \cos B\\ \sin C& \cos A& \cos B\end{array}\right| \\ \mathrm{Taking}\mathrm{out}\cos A\mathrm{and}\cos B\mathrm{common}\mathrm{from}{C}_2\mathrm{and}{C}_3,\mathrm{respectively} \\ =\cos A\cos B\left|\begin{array}{ccc}\sin A& 1& 1\\ \sin B& 1& 1\\ \sin C& 1& 1\end{array}\right| \\ =\cos A\cos B\left(0\right)\left(\because \mathrm{if}\mathrm{any}\mathrm{two}\mathrm{columns}\mathrm{are}\mathrm{identical}\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{determinant}\mathrm{is}\mathrm{zero}\right) \\ =0 \end{gathered}$$

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