Question

If $A+B+C=\pi ,$ then the value of $\left|\begin{array}{ccc}\sin (A+B+C)& \sin B& \cos C\\ -\sin B& 0& \tan A\\ \cos (A+B)& -\tan A& 0a\end{array}\right|$ 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 $\Delta =\left|\begin{array}{ccc}\sin (A+B+C)& \sin B& csoC\\ -\sin B& 0& \tan A\\ \cos (A+B)& -\tan A& 0\end{array}\right|$
$=\left|\begin{array}{ccc}\sin (\pi )& \sin B& \cos C\\ -\sin B& 0& \tan A\\ \cos (\pi -C)& -\tan A& 0\end{array}\right|$
$=\left|\begin{array}{ccc}0& \sin B& \cos C\\ -\sin B& 0& \tan A\\ \cos C& -\tan A& 0\end{array}\right|$
$=0$ ( $\because$ It is a skew-symmetric determinant of odd order)