Question

If $ A+B+C=\pi , $ then the value of $ \left| \begin{matrix} \sin \,(A+B+C) & \sin B & \cos \,C \\ -\sin \,B & 0 & \tan \,A \\ \cos \,(A+B) & -\tan \,A & 0a \\ \end{matrix} \right| $ is

• A. 0
• B. 1
• C. 2 sin B tan A cos C
• D. None of the above

Answer 1

Answer: (A)

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)