A) 0
B) 1
C) \[1+\sin \alpha \sin \beta \sin \gamma \]
D) \[1-(sin\alpha -sin\beta )(sin\beta -sin\gamma )\]\[(sin\gamma -sin\alpha )\]
Correct Answer: A
Solution :
Given, \[\left| \begin{matrix} \sin \alpha & \cos \alpha & \sin (\alpha +\delta ) \\ \sin \beta & \cos \beta & \sin (\beta +\delta ) \\ \sin \gamma & \cos \gamma & \sin (\gamma +\delta ) \\ \end{matrix} \right|\] \[=\left| \begin{matrix} \sin \alpha & \cos \alpha & \sin \alpha .\cos \delta +\cos \alpha .\sin \delta \\ \sin \beta & \cos \beta & \sin \beta .\cos \delta +\cos \beta .\sin \delta \\ \sin \gamma & \cos \gamma & \sin \gamma .\cos \delta +\cos \gamma .\sin \delta \\ \end{matrix} \right|\] \[=\left| \begin{matrix} \sin \alpha & \cos \alpha & \sin \alpha .\cos \delta \\ \sin \beta & \cos \beta & \sin \beta .\cos \delta \\ \sin \gamma & \cos \gamma & \sin \gamma .\cos \delta \\ \end{matrix} \right|\] \[+\left| \begin{matrix} \sin \alpha & \cos \alpha & \cos \alpha .\sin \delta \\ \sin \beta & \cos \beta & \cos \beta .\sin \delta \\ \sin \gamma & \cos \gamma & \cos \gamma .\sin \delta \\ \end{matrix} \right|\] \[=\cos \delta \left| \begin{matrix} \sin \alpha & \cos \alpha & \sin \alpha \\ \sin \beta & \cos \beta & \sin \beta \\ \sin \gamma & \cos \gamma & \sin \gamma \\ \end{matrix} \right|\] \[+\sin \delta \left| \begin{matrix} \sin \alpha & \cos \alpha & \cos \alpha \\ \sin \beta & \cos \beta & \cos \beta \\ \sin \gamma & \cos \gamma & \cos \gamma \\ \end{matrix} \right|\] \[=\cos \delta \times 0+\sin \delta \times 0\] ( \[\because \] \[{{C}_{1}}\] and \[{{C}_{2}}\] are identical \[{{C}_{2}}\] and \[{{C}_{3}}\] identical) \[=0\]
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