A) 0
B) 1
C) 2 sin B tan A cos C
D) None of the above
Correct Answer: A
Solution :
Given that, \[A+B+C=\pi \] Let \[\Delta =\left| \begin{matrix} \sin \,(A+B+C) & \sin \,B & cso\,C \\ -\sin \,B & 0 & \tan \,A \\ \cos \,(A+B) & -\tan \,A & 0 \\ \end{matrix} \right|\] \[=\left| \begin{matrix} \sin (\pi ) & \sin \,B & \cos \,C \\ -\sin B & 0 & \tan A \\ \cos \,(\pi -C) & -\tan \,A & 0 \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 0 & \sin \,B & \cos \,C \\ -\sin B & 0 & \tan A \\ \cos \,C & -\tan \,A & 0 \\ \end{matrix} \right|\] \[=\,\,0\] (\[\because \] It is a skew-symmetric determinant of odd order)
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