Question

If ${\cot }^{-1}x+{\cot }^{-1}y+{\cot }^{-1}z=\frac{\pi }{2}$, then $x+y+z$ is equal to

• A. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
• B. $xyz$
• C. $xy+yz+zx$
• D. $2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$

Answer 1

The correct option is B $xyz$

${\cot }^{-1}x+{\cot }^{-1}y+{\cot }^{-1}z=\frac{\pi }{2}$
$\Rightarrow \sum (\frac{\pi }{2}-{\tan }^{-1}x)=\frac{\pi }{2}\Rightarrow \sum \left({\tan }^{-1}x\right)=\pi \Rightarrow {\tan }^{-1}\left\{\frac{x+y+x-xyz}{1-(xy+yz+zx)}\right\}=\pi \Rightarrow \frac{x+y+x-xyz}{1-(xy+yz+zx)}=\tan \pi =0$
$\Rightarrow x+y+z=xyz$