Q101 / 170 · Manipal (MET) 2021 · Mathematics
. If $A+B+C=\pi$, then $\left|\begin{array}{ccc}\sin (A+B+C) & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ \cos (A+B) & -\tan A & 0\end{array}\right|$ is equal to
- A$\sin A$
- B$\sin A \cos B$
- Correct answerC0
- DNone of these
Solution
In a triangle $A B C$,
$A+B+C=\pi$,
$\Rightarrow \sin (A+B+C)=0, \cos (A+B)=-\cos C$
$\begin{aligned} \therefore \quad\left|\begin{array}{ccc}\sin (A+B+C) & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ \cos (A+B) & -\tan A & 0\end{array}\right| \\ \quad=\left|\begin{array}{ccc}0 & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ -\cos C & -\tan A & 0\end{array}\right|\end{aligned}$
=0 $\therefore \text{It is the determinant of a skew symmetric matrix of odd order. }$
=0 $\therefore \text{It is the determinant of a skew symmetric matrix of odd order. }$
Hence, the correct answer is option: 'C'.