Question

If ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\pi$ then $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=$

• A. $0$
• B. $1$
• C. $\frac{1}{xyz}$
• D. $xyz$

Answer 1

The correct option is B $1$

Given, ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\pi \Rightarrow {\tan }^{-1}x+{\tan }^{-1}y=\pi -{\tan }^{-1}z$
$\Rightarrow {\tan }^{-1}\left(\frac{x+y}{1-xy}\right)=\pi -{\tan }^{-1}z$
$\Rightarrow \left(\frac{x+y}{1-xy}\right)=\tan (\pi -{\tan }^{-1}z)$
$\Rightarrow \frac{x+y}{1-xy}=-\tan \left({\tan }^{-1}z\right)\Rightarrow \frac{x+y}{1-xy}=-z\Rightarrow x+y=-z+xyz\Rightarrow x+y+z=xyz$
$\Rightarrow \frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}=1$