Question

Let $x=\tan A,y=\tan B,z=\tan C$ and $x+y+z-xyz=0,$ then $A+B+C=$

• A. $\frac{n\pi }{3}$
• B. $\frac{n\pi }{2}$
• C. $n\pi$
• D. $\frac{n\pi }{4}$

Answer 1

The correct option is C $n\pi$

Given $x=\tan A,y=\tan B,z=\tan C$
and $x+y+z-xyz=0$
$\therefore \tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-(\tan A\tan B+\tan B\tan C+\tan C\tan A)}$
$\tan (A+B+C)=\frac{x+y+z-xyz}{1-(xy+yz+xz)}=0$

$⟹\tan (A+B+C)=0$

$⟹(A+B+C)={\tan }^{-1}\left(0\right)=n\pi$
$⟹A+B+C=n\pi$