Question

If $A+B+C=\pi$, $n\in z$, then $\tan nA+\tan nB+\tan nC$ is equal to :

• A. $0$
• B. $1$
• C. $\tan nA.\tan nB.\tan nC$
• D. none of these

Answer 1

Answer: (C)

$\tan nA.\tan nB.\tan nC$

Given $A+B+C=\pi \Rightarrow nA+nB+nC=n\pi$

$\therefore \tan (nA+nB+nC)=\frac{\tan nA+\tan nB+\tan nC-\tan nA.\tan nB.\tan nC}{1-\sum \tan nA.\tan nB}$

$=\tan n\pi =0$

$\Rightarrow \tan nA+\tan nB+\tan nC=\tan nA.\tan nB.\tan nC$