Question

If $A+B+C=\pi$, then $\left|\begin{array}{ccc}\tan (A+B+C)& \tan B& \tan C\\ \tan (A+C)& 0& \tan A\\ \tan (A+B)& -\tan A& 0\end{array}\right|$ is equal to

• A. $0$
• B. $1$
• C. $tanAtanBtanC$
• D. $-2$

Answer 1

The correct option is A $0$

Let $P=\left|\begin{array}{ccc}tan(A+B+C)& tanB& tanC\\ tan(A+C)& 0& tanA\\ tan(A+B)& -tanA& 0\end{array}\right|$

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

$=\tan (A+B+C){\tan }^{2}A+\tan A\tan B\tan (A+B)-\tan A\tan C\tan (A+C)$

Given $A+B+C=\pi ⟹A+B=\pi -C$ or $A+C=\pi -B$

$\therefore P=\tan \left(\pi \right){\tan }^{2}A+\tan A\tan B(-\tan C)-\tan A\tan C(-\tan B)$

$=0\cdot {\tan }^{2}A-\tan A\tan B\tan C+\tan A\tan B\tan C$

$=0$