Question

If ${\cot }^{-1}x+{\cot }^{-1}y+{\cot }^{-1}z=\frac{\pi }{2},x,y,z>0$ and $xy<1$, then $x+y+z$ is also equal to

• A. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
• B. $xyz$
• C. $xy+yz+zx$
• D. none of these

Answer 1

The correct option is B $xyz$

$co{t}^{-1}\left(x\right)+co{t}^{-1}\left(y\right)+co{t}^{-1}\left(z\right)=\frac{\pi }{2}$
$ta{n}^{-1}\left(\frac{1}{x}\right)+ta{n}^{-1}\left(\frac{1}{y}\right)=ta{n}^{-1}\left(z\right)$
$ta{n}^{-1}\left(\frac{x+y}{xy-1}\right)=ta{n}^{-1}\left(z\right)$
$\frac{x+y}{xy-1}=z$
$x+y=xyz-z$
$x+y+z=xyz$