Question

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. $0$
• B. $2\sin B\tan A\cos C$
• C. $1$
• D. None of these

Answer 1

Answer: (A)

$0$

Given $A+B+C=\pi$

$=\left|\begin{array}{ccc}\sin (A+B+C)& \sin B& \cos C\\ -\sin B& 0& \tan A\\ \cos A+B& -\tan A& 0\end{array}\right|$

$=\left|\begin{array}{ccc}\sin \pi & \sin B& \cos C\\ -\sin B& 0& \tan A\\ \cos (\pi -C)& -\tan A& 0\end{array}\right|$

$=\sin \pi \left[{\tan }^{2}A\right]-\sin B[-\tan A\cos (\pi -C\left)\right]+\cos C\left[\sin B\tan A\right]$

$=0+\sin B\tan A\cos (\pi -C)+\cos C\sin B\tan A$

$=\sin B\tan A[\cos \pi \cos C+\sin \pi \sin C]+\cos C\sin B\tan A$

$=-\sin B\tan A\cos C+\cos C\sin B\tan A$

$=0$ $[\therefore \sin \pi =0,\cos \pi =-1]$