Question

If $A+B+C={180}^o$ and $\tan 3A+\tan 3B+\tan 3C=k\tan 3A.\tan 3B.\tan 3C$ then $k$=

• A. $1$
• B. $3$
• C. $2$
• D. $4$

Answer 1

The correct option is A $1$

$\begin{array}{c}A+B+C=180\\ \tan 3A+\tan 3B+\tan 3C=K\tan 3A\tan 3B\tan 3C\\ A+B=180-C\\ \tan \left(3(A+B\right)=\tan \left(3\left(180-C\right)\right)\\ \frac{\tan 3A+\tan 3B}{1-\tan 3A+\tan 3B}=-\tan 3C\\ \tan 3A+\tan 3B=-\tan 3C+\tan 3A\tan 3B\tan 3C\\ \tan 3A+\tan 3B+\tan 3C=\tan 3A\tan 3B\tan 3C\end{array}$

$k=1$